Moles of H 2 O produced from 15.0 mL of 0.250 M HCl have to be determined. Concept Introduction: Stoichiometry describes quantitative relationships between reactants and products in any chemical reactions. In these problems, amount of one species is determined with help of known amount of another species.

Question

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Answer 1

Question

Chapter 14, Problem 37PE

(a)

Interpretation Introduction

Interpretation:

Moles of $H2O$ produced from $15.0 mL$ of $0.250 M HCl$ have to be determined.

Concept Introduction:

Stoichiometry describes quantitative relationships between reactants and products in any chemical reactions. In these problems, amount of one species is determined with help of known amount of another species.

(a)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

Expression for molarity of $HCl$ is as follows:

$Molarity of HCl=Moles of HClVolume(L) of HCl$ (1)

Rearrange equation (1) for moles of $HCl$ .

$Moles of HCl=(Molarity of HCl)(Volume of HCl)$ (2)

Substitute $15.0 mL$ for volume and $0.250 M$ for molarity of $HCl$ in equation (2).

$Moles of HCl=(0.250 M)(15.0 mL)(10−3 L1 mL)=0.00375 mol$

According to balanced chemical equation, two moles of $KMnO4$ react with sixteen moles of $HCl$ and produce two moles of $MnCl2$ , eight moles of $H2O$ and two moles of $KCl$ . Moles of $H2O$ produced from 0.00375 moles of $HCl$ is calculated as follows:

$Moles of H2O=(8 mol H2O16 mol HCl)(0.00375 mol HCl)=0.001875 mol$

Hence, amount of $H2O$ is $0.001875 mol$ .

(b)

Interpretation Introduction

Interpretation:

Volume of $0.150 M KMnO4$ produced from $1.85 mol MnCl2$ have to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

According to balanced chemical equation, two moles of $KMnO4$ react with sixteen moles of $HCl$ and produce two moles of $MnCl2$ , eight moles of $H2O$ and two moles of $KCl$ . Amount of $KMnO4$ produced by 1.85 moles of $MnCl2$ is calculated as follows:

$Moles of KMnO4=(2 mol KMnO42 mol MnCl2)(1.85 mol MnCl2)=1.85 mol$

Expression for molarity of $KMnO4$ is as follows:

$Molarity of KMnO4=Moles of KMnO4Volume(L) of KMnO4$ (3)

Rearrange equation (3) for volume of $KMnO4$ .

$Volume of KMnO4=Moles of KMnO4Molarity of KMnO4$ (4)

Substitute $1.85 mol$ for moles and $0.150 M$ for molarity of $KMnO4$ in equation (4).

$Volume of KMnO4=1.85 mol0.150 M=12.33 L$

Hence, volume of $KMnO4$ is $12.33 L$ .

(c)

Interpretation Introduction

Interpretation:

Volume of $2.50 M HCl$ required to produce $125 mL$ of $0.525 M KCl$ have to be determined.

Concept Introduction:

Expression for molarity equation is as follows:

$M1V1=M2V2$ (5)

Here,

$M1$ is molarity of one solution.

$V1$ is volume of one solution.

$M2$ is molarity of another solution.

$V2$ is volume of another solution.

(c)

Expert Solution

Explanation of Solution

Rearrange equation (5) for $V2$ .

$V2=M1V1M2$ (6)

Substitute $0.525 M$ for $M1$ , $125 mL$ for $V1$ and $2.50 M$ for $M2$ in equation (6) for volume of $HCl$ .

$Volume of HCl=(0.525 M)(125 mL)2.50 M=26.25 mL$

Hence volume of $HCl$ required is $26.25 mL$ .

(d)

Interpretation Introduction

Interpretation:

Molarity of $HCl$ solution if its $22.20 mL$ reacts with $15.60 mL$ of $0.250 M KMnO4$ has to be determined.

Concept Introduction:

Refer to part (c).

(d)

Expert Solution

Explanation of Solution

Rearrange equation (5) for $M2$ .

$M2=M1V1V2$ (7)

Substitute $0.250 M$ for $M1$ , $15.60 mL$ for $V1$ and $22.20 mL$ for $V2$ in equation (7) for molarity of $HCl$ .

$Molarity of HCl=(0.250 M)(15.60 mL)22.20 mL=0.176 M$

Hence molarity of $HCl$ required is $0.176 M$ .

(e)

Interpretation Introduction

Interpretation:

Volume $(L)$ of $Cl2$ gas at STP produced by reaction of $125 mL$ of $2.5 M HCl$ has to be determined.

Concept Introduction:

Expression for ideal gas equation is as follows:

$PV=nRT$ (8)

Here,

$P$ is pressure of gas.

$V$ is volume of gas.

$n$ is moles of gas.

$R$ is universal gas constant.

$T$ is absolute temperature of gas.

(e)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

Expression for molarity of $HCl$ is as follows:

$Molarity of HCl=Moles of HClVolume(L) of HCl solution$ (9)

Rearrange equation (9) for moles of $HCl$ .

$Moles of HCl=[(Molarity of HCl)(Volume of HCl solution)]$ (10)

Substitute $2.5 M$ for molarity and $125 mL$ for volume of $HCl$ solution in equation (10).

$Moles of HCl=(2.5 M)(125 mL)(10−3 L1 mL)=0.3125 mol$

Since 16 moles of $HCl$ produce 5 moles of $Cl2$ , amount of $Cl2$ produced by 0.3125 moles of $HCl$ is calculated as follows:

$Amount of Cl2=(5 mol Cl216 mol HCl)(0.3125 mol HCl)=0.0976 mol$

Rearrange equation (8) for $V$ .

$V=nRTP$ (11)

Substitute $273 K$ for $T$ , $0.0976 mol$ for $n$ , $0.082057 L⋅atm⋅K−1⋅mol−1$ for $R$ and $1 atm$ for $P$ in equation (11) for volume of $Cl2$ .

$Volume of Cl2=(0.0976 mol)(0.082057 L⋅atm⋅K−1⋅mol−1)(273 K)1 atm=2.186 L$

Hence, volume of $Cl2$ is $2.186 L$ .

(f)

Interpretation Introduction

Interpretation:

Volume $(L)$ of $Cl2$ gas at STP produced by reaction of $15.0 mL$ of $0.750 M HCl$ and $12.0 mL$ of $0.550 M KMnO4$ has to be determined.

Concept Introduction:

Refer to part (e).

(f)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

Expression for molarity of $HCl$ is as follows:

$Molarity of HCl=Moles of HClVolume(L) of HCl solution$ (9)

Rearrange equation (9) for moles of $HCl$ .

$Moles of HCl=[(Molarity of HCl)(Volume of HCl solution)]$ (10)

Substitute $0.750 M$ for molarity and $15.0 mL$ for volume of $HCl$ solution in equation (10).

$Moles of HCl=(0.750 M)(15.0 mL)(10−3 L1 mL)=0.01125 mol$

Expression for molarity of $KMnO4$ is as follows:

$Molarity of KMnO4=Moles of KMnO4Volume(L) of KMnO4$ (3)

Rearrange equation (3) for moles of $KMnO4$ .

$Moles of KMnO4=[(Molarity of KMnO4)(Volume of KMnO4)]$ (12)

Substitute $12.0 mL$ for molarity and $0.550 M$ for molarity of $KMnO4$ in equation (12).

$Moles of KMnO4=(0.550 M)(12.0 mL)(10−3 L1 mL)=0.0066 mol$

Since 16 moles of $HCl$ form 5 moles of $Cl2$ , amount of $Cl2$ produced by 0.01125 moles of $HCl$ is calculated as follows:

$Amount of Cl2=(5 mol Cl216 mol HCl)(0.01125 mol HCl)=0.003516 mol$

Since 2 moles of $KMnO4$ form 5 moles of $Cl2$ , amount of $Cl2$ produced by 0.0066 moles of $KMnO4$ is calculated as follows:

$Amount of Cl2=(5 mol Cl22 mol KMnO4)(0.0066 mol KMnO4)=0.0165 mol$

Since $HCl$ produces less amount of $Cl2$ than $KMnO4$ , $HCl$ acts as limiting reactant and amount of $Cl2$ will be produced in accordance with its amount only.

Rearrange equation (8) for $V$ .

$V=nRTP$ (11)

Substitute $273 K$ for $T$ , $0.003516 mol$ for $n$ , $0.082057 L⋅atm⋅K−1⋅mol−1$ for $R$ and $1 atm$ for $P$ in equation (11) for volume of $Cl2$ .

$Volume of Cl2=(0.003516 mol)(0.082057 L⋅atm⋅K−1⋅mol−1)(273 K)1 atm=0.079 L$

Hence, volume of $Cl2$ is $0.079 L$ .

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Answer 2

Question

Chapter 14, Problem 37PE

(a)

Interpretation Introduction

Interpretation:

Moles of $H2O$ produced from $15.0 mL$ of $0.250 M HCl$ have to be determined.

Concept Introduction:

Stoichiometry describes quantitative relationships between reactants and products in any chemical reactions. In these problems, amount of one species is determined with help of known amount of another species.

(a)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

Expression for molarity of $HCl$ is as follows:

$Molarity of HCl=Moles of HClVolume(L) of HCl$ (1)

Rearrange equation (1) for moles of $HCl$ .

$Moles of HCl=(Molarity of HCl)(Volume of HCl)$ (2)

Substitute $15.0 mL$ for volume and $0.250 M$ for molarity of $HCl$ in equation (2).

$Moles of HCl=(0.250 M)(15.0 mL)(10−3 L1 mL)=0.00375 mol$

According to balanced chemical equation, two moles of $KMnO4$ react with sixteen moles of $HCl$ and produce two moles of $MnCl2$ , eight moles of $H2O$ and two moles of $KCl$ . Moles of $H2O$ produced from 0.00375 moles of $HCl$ is calculated as follows:

$Moles of H2O=(8 mol H2O16 mol HCl)(0.00375 mol HCl)=0.001875 mol$

Hence, amount of $H2O$ is $0.001875 mol$ .

(b)

Interpretation Introduction

Interpretation:

Volume of $0.150 M KMnO4$ produced from $1.85 mol MnCl2$ have to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

According to balanced chemical equation, two moles of $KMnO4$ react with sixteen moles of $HCl$ and produce two moles of $MnCl2$ , eight moles of $H2O$ and two moles of $KCl$ . Amount of $KMnO4$ produced by 1.85 moles of $MnCl2$ is calculated as follows:

$Moles of KMnO4=(2 mol KMnO42 mol MnCl2)(1.85 mol MnCl2)=1.85 mol$

Expression for molarity of $KMnO4$ is as follows:

$Molarity of KMnO4=Moles of KMnO4Volume(L) of KMnO4$ (3)

Rearrange equation (3) for volume of $KMnO4$ .

$Volume of KMnO4=Moles of KMnO4Molarity of KMnO4$ (4)

Substitute $1.85 mol$ for moles and $0.150 M$ for molarity of $KMnO4$ in equation (4).

$Volume of KMnO4=1.85 mol0.150 M=12.33 L$

Hence, volume of $KMnO4$ is $12.33 L$ .

(c)

Interpretation Introduction

Interpretation:

Volume of $2.50 M HCl$ required to produce $125 mL$ of $0.525 M KCl$ have to be determined.

Concept Introduction:

Expression for molarity equation is as follows:

$M1V1=M2V2$ (5)

Here,

$M1$ is molarity of one solution.

$V1$ is volume of one solution.

$M2$ is molarity of another solution.

$V2$ is volume of another solution.

(c)

Expert Solution

Explanation of Solution

Rearrange equation (5) for $V2$ .

$V2=M1V1M2$ (6)

Substitute $0.525 M$ for $M1$ , $125 mL$ for $V1$ and $2.50 M$ for $M2$ in equation (6) for volume of $HCl$ .

$Volume of HCl=(0.525 M)(125 mL)2.50 M=26.25 mL$

Hence volume of $HCl$ required is $26.25 mL$ .

(d)

Interpretation Introduction

Interpretation:

Molarity of $HCl$ solution if its $22.20 mL$ reacts with $15.60 mL$ of $0.250 M KMnO4$ has to be determined.

Concept Introduction:

Refer to part (c).

(d)

Expert Solution

Explanation of Solution

Rearrange equation (5) for $M2$ .

$M2=M1V1V2$ (7)

Substitute $0.250 M$ for $M1$ , $15.60 mL$ for $V1$ and $22.20 mL$ for $V2$ in equation (7) for molarity of $HCl$ .

$Molarity of HCl=(0.250 M)(15.60 mL)22.20 mL=0.176 M$

Hence molarity of $HCl$ required is $0.176 M$ .

(e)

Interpretation Introduction

Interpretation:

Volume $(L)$ of $Cl2$ gas at STP produced by reaction of $125 mL$ of $2.5 M HCl$ has to be determined.

Concept Introduction:

Expression for ideal gas equation is as follows:

$PV=nRT$ (8)

Here,

$P$ is pressure of gas.

$V$ is volume of gas.

$n$ is moles of gas.

$R$ is universal gas constant.

$T$ is absolute temperature of gas.

(e)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

Expression for molarity of $HCl$ is as follows:

$Molarity of HCl=Moles of HClVolume(L) of HCl solution$ (9)

Rearrange equation (9) for moles of $HCl$ .

$Moles of HCl=[(Molarity of HCl)(Volume of HCl solution)]$ (10)

Substitute $2.5 M$ for molarity and $125 mL$ for volume of $HCl$ solution in equation (10).

$Moles of HCl=(2.5 M)(125 mL)(10−3 L1 mL)=0.3125 mol$

Since 16 moles of $HCl$ produce 5 moles of $Cl2$ , amount of $Cl2$ produced by 0.3125 moles of $HCl$ is calculated as follows:

$Amount of Cl2=(5 mol Cl216 mol HCl)(0.3125 mol HCl)=0.0976 mol$

Rearrange equation (8) for $V$ .

$V=nRTP$ (11)

Substitute $273 K$ for $T$ , $0.0976 mol$ for $n$ , $0.082057 L⋅atm⋅K−1⋅mol−1$ for $R$ and $1 atm$ for $P$ in equation (11) for volume of $Cl2$ .

$Volume of Cl2=(0.0976 mol)(0.082057 L⋅atm⋅K−1⋅mol−1)(273 K)1 atm=2.186 L$

Hence, volume of $Cl2$ is $2.186 L$ .

(f)

Interpretation Introduction

Interpretation:

Volume $(L)$ of $Cl2$ gas at STP produced by reaction of $15.0 mL$ of $0.750 M HCl$ and $12.0 mL$ of $0.550 M KMnO4$ has to be determined.

Concept Introduction:

Refer to part (e).

(f)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

Expression for molarity of $HCl$ is as follows:

$Molarity of HCl=Moles of HClVolume(L) of HCl solution$ (9)

Rearrange equation (9) for moles of $HCl$ .

$Moles of HCl=[(Molarity of HCl)(Volume of HCl solution)]$ (10)

Substitute $0.750 M$ for molarity and $15.0 mL$ for volume of $HCl$ solution in equation (10).

$Moles of HCl=(0.750 M)(15.0 mL)(10−3 L1 mL)=0.01125 mol$

Expression for molarity of $KMnO4$ is as follows:

$Molarity of KMnO4=Moles of KMnO4Volume(L) of KMnO4$ (3)

Rearrange equation (3) for moles of $KMnO4$ .

$Moles of KMnO4=[(Molarity of KMnO4)(Volume of KMnO4)]$ (12)

Substitute $12.0 mL$ for molarity and $0.550 M$ for molarity of $KMnO4$ in equation (12).

$Moles of KMnO4=(0.550 M)(12.0 mL)(10−3 L1 mL)=0.0066 mol$

Since 16 moles of $HCl$ form 5 moles of $Cl2$ , amount of $Cl2$ produced by 0.01125 moles of $HCl$ is calculated as follows:

$Amount of Cl2=(5 mol Cl216 mol HCl)(0.01125 mol HCl)=0.003516 mol$

Since 2 moles of $KMnO4$ form 5 moles of $Cl2$ , amount of $Cl2$ produced by 0.0066 moles of $KMnO4$ is calculated as follows:

$Amount of Cl2=(5 mol Cl22 mol KMnO4)(0.0066 mol KMnO4)=0.0165 mol$

Since $HCl$ produces less amount of $Cl2$ than $KMnO4$ , $HCl$ acts as limiting reactant and amount of $Cl2$ will be produced in accordance with its amount only.

Rearrange equation (8) for $V$ .

$V=nRTP$ (11)

Substitute $273 K$ for $T$ , $0.003516 mol$ for $n$ , $0.082057 L⋅atm⋅K−1⋅mol−1$ for $R$ and $1 atm$ for $P$ in equation (11) for volume of $Cl2$ .

$Volume of Cl2=(0.003516 mol)(0.082057 L⋅atm⋅K−1⋅mol−1)(273 K)1 atm=0.079 L$

Hence, volume of $Cl2$ is $0.079 L$ .

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Answer 3

Question

Chapter 14, Problem 37PE

(a)

Interpretation Introduction

Interpretation:

Moles of $H2O$ produced from $15.0 mL$ of $0.250 M HCl$ have to be determined.

Concept Introduction:

Stoichiometry describes quantitative relationships between reactants and products in any chemical reactions. In these problems, amount of one species is determined with help of known amount of another species.

(a)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

Expression for molarity of $HCl$ is as follows:

$Molarity of HCl=Moles of HClVolume(L) of HCl$ (1)

Rearrange equation (1) for moles of $HCl$ .

$Moles of HCl=(Molarity of HCl)(Volume of HCl)$ (2)

Substitute $15.0 mL$ for volume and $0.250 M$ for molarity of $HCl$ in equation (2).

$Moles of HCl=(0.250 M)(15.0 mL)(10−3 L1 mL)=0.00375 mol$

According to balanced chemical equation, two moles of $KMnO4$ react with sixteen moles of $HCl$ and produce two moles of $MnCl2$ , eight moles of $H2O$ and two moles of $KCl$ . Moles of $H2O$ produced from 0.00375 moles of $HCl$ is calculated as follows:

$Moles of H2O=(8 mol H2O16 mol HCl)(0.00375 mol HCl)=0.001875 mol$

Hence, amount of $H2O$ is $0.001875 mol$ .

(b)

Interpretation Introduction

Interpretation:

Volume of $0.150 M KMnO4$ produced from $1.85 mol MnCl2$ have to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

According to balanced chemical equation, two moles of $KMnO4$ react with sixteen moles of $HCl$ and produce two moles of $MnCl2$ , eight moles of $H2O$ and two moles of $KCl$ . Amount of $KMnO4$ produced by 1.85 moles of $MnCl2$ is calculated as follows:

$Moles of KMnO4=(2 mol KMnO42 mol MnCl2)(1.85 mol MnCl2)=1.85 mol$

Expression for molarity of $KMnO4$ is as follows:

$Molarity of KMnO4=Moles of KMnO4Volume(L) of KMnO4$ (3)

Rearrange equation (3) for volume of $KMnO4$ .

$Volume of KMnO4=Moles of KMnO4Molarity of KMnO4$ (4)

Substitute $1.85 mol$ for moles and $0.150 M$ for molarity of $KMnO4$ in equation (4).

$Volume of KMnO4=1.85 mol0.150 M=12.33 L$

Hence, volume of $KMnO4$ is $12.33 L$ .

(c)

Interpretation Introduction

Interpretation:

Volume of $2.50 M HCl$ required to produce $125 mL$ of $0.525 M KCl$ have to be determined.

Concept Introduction:

Expression for molarity equation is as follows:

$M1V1=M2V2$ (5)

Here,

$M1$ is molarity of one solution.

$V1$ is volume of one solution.

$M2$ is molarity of another solution.

$V2$ is volume of another solution.

(c)

Expert Solution

Explanation of Solution

Rearrange equation (5) for $V2$ .

$V2=M1V1M2$ (6)

Substitute $0.525 M$ for $M1$ , $125 mL$ for $V1$ and $2.50 M$ for $M2$ in equation (6) for volume of $HCl$ .

$Volume of HCl=(0.525 M)(125 mL)2.50 M=26.25 mL$

Hence volume of $HCl$ required is $26.25 mL$ .

(d)

Interpretation Introduction

Interpretation:

Molarity of $HCl$ solution if its $22.20 mL$ reacts with $15.60 mL$ of $0.250 M KMnO4$ has to be determined.

Concept Introduction:

Refer to part (c).

(d)

Expert Solution

Explanation of Solution

Rearrange equation (5) for $M2$ .

$M2=M1V1V2$ (7)

Substitute $0.250 M$ for $M1$ , $15.60 mL$ for $V1$ and $22.20 mL$ for $V2$ in equation (7) for molarity of $HCl$ .

$Molarity of HCl=(0.250 M)(15.60 mL)22.20 mL=0.176 M$

Hence molarity of $HCl$ required is $0.176 M$ .

(e)

Interpretation Introduction

Interpretation:

Volume $(L)$ of $Cl2$ gas at STP produced by reaction of $125 mL$ of $2.5 M HCl$ has to be determined.

Concept Introduction:

Expression for ideal gas equation is as follows:

$PV=nRT$ (8)

Here,

$P$ is pressure of gas.

$V$ is volume of gas.

$n$ is moles of gas.

$R$ is universal gas constant.

$T$ is absolute temperature of gas.

(e)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

Expression for molarity of $HCl$ is as follows:

$Molarity of HCl=Moles of HClVolume(L) of HCl solution$ (9)

Rearrange equation (9) for moles of $HCl$ .

$Moles of HCl=[(Molarity of HCl)(Volume of HCl solution)]$ (10)

Substitute $2.5 M$ for molarity and $125 mL$ for volume of $HCl$ solution in equation (10).

$Moles of HCl=(2.5 M)(125 mL)(10−3 L1 mL)=0.3125 mol$

Since 16 moles of $HCl$ produce 5 moles of $Cl2$ , amount of $Cl2$ produced by 0.3125 moles of $HCl$ is calculated as follows:

$Amount of Cl2=(5 mol Cl216 mol HCl)(0.3125 mol HCl)=0.0976 mol$

Rearrange equation (8) for $V$ .

$V=nRTP$ (11)

Substitute $273 K$ for $T$ , $0.0976 mol$ for $n$ , $0.082057 L⋅atm⋅K−1⋅mol−1$ for $R$ and $1 atm$ for $P$ in equation (11) for volume of $Cl2$ .

$Volume of Cl2=(0.0976 mol)(0.082057 L⋅atm⋅K−1⋅mol−1)(273 K)1 atm=2.186 L$

Hence, volume of $Cl2$ is $2.186 L$ .

(f)

Interpretation Introduction

Interpretation:

Volume $(L)$ of $Cl2$ gas at STP produced by reaction of $15.0 mL$ of $0.750 M HCl$ and $12.0 mL$ of $0.550 M KMnO4$ has to be determined.

Concept Introduction:

Refer to part (e).

(f)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

Expression for molarity of $HCl$ is as follows:

$Molarity of HCl=Moles of HClVolume(L) of HCl solution$ (9)

Rearrange equation (9) for moles of $HCl$ .

$Moles of HCl=[(Molarity of HCl)(Volume of HCl solution)]$ (10)

Substitute $0.750 M$ for molarity and $15.0 mL$ for volume of $HCl$ solution in equation (10).

$Moles of HCl=(0.750 M)(15.0 mL)(10−3 L1 mL)=0.01125 mol$

Expression for molarity of $KMnO4$ is as follows:

$Molarity of KMnO4=Moles of KMnO4Volume(L) of KMnO4$ (3)

Rearrange equation (3) for moles of $KMnO4$ .

$Moles of KMnO4=[(Molarity of KMnO4)(Volume of KMnO4)]$ (12)

Substitute $12.0 mL$ for molarity and $0.550 M$ for molarity of $KMnO4$ in equation (12).

$Moles of KMnO4=(0.550 M)(12.0 mL)(10−3 L1 mL)=0.0066 mol$

Since 16 moles of $HCl$ form 5 moles of $Cl2$ , amount of $Cl2$ produced by 0.01125 moles of $HCl$ is calculated as follows:

$Amount of Cl2=(5 mol Cl216 mol HCl)(0.01125 mol HCl)=0.003516 mol$

Since 2 moles of $KMnO4$ form 5 moles of $Cl2$ , amount of $Cl2$ produced by 0.0066 moles of $KMnO4$ is calculated as follows:

$Amount of Cl2=(5 mol Cl22 mol KMnO4)(0.0066 mol KMnO4)=0.0165 mol$

Since $HCl$ produces less amount of $Cl2$ than $KMnO4$ , $HCl$ acts as limiting reactant and amount of $Cl2$ will be produced in accordance with its amount only.

Rearrange equation (8) for $V$ .

$V=nRTP$ (11)

Substitute $273 K$ for $T$ , $0.003516 mol$ for $n$ , $0.082057 L⋅atm⋅K−1⋅mol−1$ for $R$ and $1 atm$ for $P$ in equation (11) for volume of $Cl2$ .

$Volume of Cl2=(0.003516 mol)(0.082057 L⋅atm⋅K−1⋅mol−1)(273 K)1 atm=0.079 L$

Hence, volume of $Cl2$ is $0.079 L$ .

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Answer 4

Question

Chapter 14, Problem 37PE

(a)

Interpretation Introduction

Interpretation:

Moles of $H2O$ produced from $15.0 mL$ of $0.250 M HCl$ have to be determined.

Concept Introduction:

Stoichiometry describes quantitative relationships between reactants and products in any chemical reactions. In these problems, amount of one species is determined with help of known amount of another species.

(a)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

Expression for molarity of $HCl$ is as follows:

$Molarity of HCl=Moles of HClVolume(L) of HCl$ (1)

Rearrange equation (1) for moles of $HCl$ .

$Moles of HCl=(Molarity of HCl)(Volume of HCl)$ (2)

Substitute $15.0 mL$ for volume and $0.250 M$ for molarity of $HCl$ in equation (2).

$Moles of HCl=(0.250 M)(15.0 mL)(10−3 L1 mL)=0.00375 mol$

According to balanced chemical equation, two moles of $KMnO4$ react with sixteen moles of $HCl$ and produce two moles of $MnCl2$ , eight moles of $H2O$ and two moles of $KCl$ . Moles of $H2O$ produced from 0.00375 moles of $HCl$ is calculated as follows:

$Moles of H2O=(8 mol H2O16 mol HCl)(0.00375 mol HCl)=0.001875 mol$

Hence, amount of $H2O$ is $0.001875 mol$ .

(b)

Interpretation Introduction

Interpretation:

Volume of $0.150 M KMnO4$ produced from $1.85 mol MnCl2$ have to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

According to balanced chemical equation, two moles of $KMnO4$ react with sixteen moles of $HCl$ and produce two moles of $MnCl2$ , eight moles of $H2O$ and two moles of $KCl$ . Amount of $KMnO4$ produced by 1.85 moles of $MnCl2$ is calculated as follows:

$Moles of KMnO4=(2 mol KMnO42 mol MnCl2)(1.85 mol MnCl2)=1.85 mol$

Expression for molarity of $KMnO4$ is as follows:

$Molarity of KMnO4=Moles of KMnO4Volume(L) of KMnO4$ (3)

Rearrange equation (3) for volume of $KMnO4$ .

$Volume of KMnO4=Moles of KMnO4Molarity of KMnO4$ (4)

Substitute $1.85 mol$ for moles and $0.150 M$ for molarity of $KMnO4$ in equation (4).

$Volume of KMnO4=1.85 mol0.150 M=12.33 L$

Hence, volume of $KMnO4$ is $12.33 L$ .

(c)

Interpretation Introduction

Interpretation:

Volume of $2.50 M HCl$ required to produce $125 mL$ of $0.525 M KCl$ have to be determined.

Concept Introduction:

Expression for molarity equation is as follows:

$M1V1=M2V2$ (5)

Here,

$M1$ is molarity of one solution.

$V1$ is volume of one solution.

$M2$ is molarity of another solution.

$V2$ is volume of another solution.

(c)

Expert Solution

Explanation of Solution

Rearrange equation (5) for $V2$ .

$V2=M1V1M2$ (6)

Substitute $0.525 M$ for $M1$ , $125 mL$ for $V1$ and $2.50 M$ for $M2$ in equation (6) for volume of $HCl$ .

$Volume of HCl=(0.525 M)(125 mL)2.50 M=26.25 mL$

Hence volume of $HCl$ required is $26.25 mL$ .

(d)

Interpretation Introduction

Interpretation:

Molarity of $HCl$ solution if its $22.20 mL$ reacts with $15.60 mL$ of $0.250 M KMnO4$ has to be determined.

Concept Introduction:

Refer to part (c).

(d)

Expert Solution

Explanation of Solution

Rearrange equation (5) for $M2$ .

$M2=M1V1V2$ (7)

Substitute $0.250 M$ for $M1$ , $15.60 mL$ for $V1$ and $22.20 mL$ for $V2$ in equation (7) for molarity of $HCl$ .

$Molarity of HCl=(0.250 M)(15.60 mL)22.20 mL=0.176 M$

Hence molarity of $HCl$ required is $0.176 M$ .

(e)

Interpretation Introduction

Interpretation:

Volume $(L)$ of $Cl2$ gas at STP produced by reaction of $125 mL$ of $2.5 M HCl$ has to be determined.

Concept Introduction:

Expression for ideal gas equation is as follows:

$PV=nRT$ (8)

Here,

$P$ is pressure of gas.

$V$ is volume of gas.

$n$ is moles of gas.

$R$ is universal gas constant.

$T$ is absolute temperature of gas.

(e)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

Expression for molarity of $HCl$ is as follows:

$Molarity of HCl=Moles of HClVolume(L) of HCl solution$ (9)

Rearrange equation (9) for moles of $HCl$ .

$Moles of HCl=[(Molarity of HCl)(Volume of HCl solution)]$ (10)

Substitute $2.5 M$ for molarity and $125 mL$ for volume of $HCl$ solution in equation (10).

$Moles of HCl=(2.5 M)(125 mL)(10−3 L1 mL)=0.3125 mol$

Since 16 moles of $HCl$ produce 5 moles of $Cl2$ , amount of $Cl2$ produced by 0.3125 moles of $HCl$ is calculated as follows:

$Amount of Cl2=(5 mol Cl216 mol HCl)(0.3125 mol HCl)=0.0976 mol$

Rearrange equation (8) for $V$ .

$V=nRTP$ (11)

Substitute $273 K$ for $T$ , $0.0976 mol$ for $n$ , $0.082057 L⋅atm⋅K−1⋅mol−1$ for $R$ and $1 atm$ for $P$ in equation (11) for volume of $Cl2$ .

$Volume of Cl2=(0.0976 mol)(0.082057 L⋅atm⋅K−1⋅mol−1)(273 K)1 atm=2.186 L$

Hence, volume of $Cl2$ is $2.186 L$ .

(f)

Interpretation Introduction

Interpretation:

Volume $(L)$ of $Cl2$ gas at STP produced by reaction of $15.0 mL$ of $0.750 M HCl$ and $12.0 mL$ of $0.550 M KMnO4$ has to be determined.

Concept Introduction:

Refer to part (e).

(f)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

Expression for molarity of $HCl$ is as follows:

$Molarity of HCl=Moles of HClVolume(L) of HCl solution$ (9)

Rearrange equation (9) for moles of $HCl$ .

$Moles of HCl=[(Molarity of HCl)(Volume of HCl solution)]$ (10)

Substitute $0.750 M$ for molarity and $15.0 mL$ for volume of $HCl$ solution in equation (10).

$Moles of HCl=(0.750 M)(15.0 mL)(10−3 L1 mL)=0.01125 mol$

Expression for molarity of $KMnO4$ is as follows:

$Molarity of KMnO4=Moles of KMnO4Volume(L) of KMnO4$ (3)

Rearrange equation (3) for moles of $KMnO4$ .

$Moles of KMnO4=[(Molarity of KMnO4)(Volume of KMnO4)]$ (12)

Substitute $12.0 mL$ for molarity and $0.550 M$ for molarity of $KMnO4$ in equation (12).

$Moles of KMnO4=(0.550 M)(12.0 mL)(10−3 L1 mL)=0.0066 mol$

Since 16 moles of $HCl$ form 5 moles of $Cl2$ , amount of $Cl2$ produced by 0.01125 moles of $HCl$ is calculated as follows:

$Amount of Cl2=(5 mol Cl216 mol HCl)(0.01125 mol HCl)=0.003516 mol$

Since 2 moles of $KMnO4$ form 5 moles of $Cl2$ , amount of $Cl2$ produced by 0.0066 moles of $KMnO4$ is calculated as follows:

$Amount of Cl2=(5 mol Cl22 mol KMnO4)(0.0066 mol KMnO4)=0.0165 mol$

Since $HCl$ produces less amount of $Cl2$ than $KMnO4$ , $HCl$ acts as limiting reactant and amount of $Cl2$ will be produced in accordance with its amount only.

Rearrange equation (8) for $V$ .

$V=nRTP$ (11)

Substitute $273 K$ for $T$ , $0.003516 mol$ for $n$ , $0.082057 L⋅atm⋅K−1⋅mol−1$ for $R$ and $1 atm$ for $P$ in equation (11) for volume of $Cl2$ .

$Volume of Cl2=(0.003516 mol)(0.082057 L⋅atm⋅K−1⋅mol−1)(273 K)1 atm=0.079 L$

Hence, volume of $Cl2$ is $0.079 L$ .

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Answer 5

Chapter 14, Problem 37PE

(a)

Interpretation Introduction

Interpretation:

Moles of $H2O$ produced from $15.0 mL$ of $0.250 M HCl$ have to be determined.

Concept Introduction:

Stoichiometry describes quantitative relationships between reactants and products in any chemical reactions. In these problems, amount of one species is determined with help of known amount of another species.

(a)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

Expression for molarity of $HCl$ is as follows:

$Molarity of HCl=Moles of HClVolume(L) of HCl$ (1)

Rearrange equation (1) for moles of $HCl$ .

$Moles of HCl=(Molarity of HCl)(Volume of HCl)$ (2)

Substitute $15.0 mL$ for volume and $0.250 M$ for molarity of $HCl$ in equation (2).

$Moles of HCl=(0.250 M)(15.0 mL)(10−3 L1 mL)=0.00375 mol$

According to balanced chemical equation, two moles of $KMnO4$ react with sixteen moles of $HCl$ and produce two moles of $MnCl2$ , eight moles of $H2O$ and two moles of $KCl$ . Moles of $H2O$ produced from 0.00375 moles of $HCl$ is calculated as follows:

$Moles of H2O=(8 mol H2O16 mol HCl)(0.00375 mol HCl)=0.001875 mol$

Hence, amount of $H2O$ is $0.001875 mol$ .

(b)

Interpretation Introduction

Interpretation:

Volume of $0.150 M KMnO4$ produced from $1.85 mol MnCl2$ have to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

According to balanced chemical equation, two moles of $KMnO4$ react with sixteen moles of $HCl$ and produce two moles of $MnCl2$ , eight moles of $H2O$ and two moles of $KCl$ . Amount of $KMnO4$ produced by 1.85 moles of $MnCl2$ is calculated as follows:

$Moles of KMnO4=(2 mol KMnO42 mol MnCl2)(1.85 mol MnCl2)=1.85 mol$

Expression for molarity of $KMnO4$ is as follows:

$Molarity of KMnO4=Moles of KMnO4Volume(L) of KMnO4$ (3)

Rearrange equation (3) for volume of $KMnO4$ .

$Volume of KMnO4=Moles of KMnO4Molarity of KMnO4$ (4)

Substitute $1.85 mol$ for moles and $0.150 M$ for molarity of $KMnO4$ in equation (4).

$Volume of KMnO4=1.85 mol0.150 M=12.33 L$

Hence, volume of $KMnO4$ is $12.33 L$ .

(c)

Interpretation Introduction

Interpretation:

Volume of $2.50 M HCl$ required to produce $125 mL$ of $0.525 M KCl$ have to be determined.

Concept Introduction:

Expression for molarity equation is as follows:

$M1V1=M2V2$ (5)

Here,

$M1$ is molarity of one solution.

$V1$ is volume of one solution.

$M2$ is molarity of another solution.

$V2$ is volume of another solution.

(c)

Expert Solution

Explanation of Solution

Rearrange equation (5) for $V2$ .

$V2=M1V1M2$ (6)

Substitute $0.525 M$ for $M1$ , $125 mL$ for $V1$ and $2.50 M$ for $M2$ in equation (6) for volume of $HCl$ .

$Volume of HCl=(0.525 M)(125 mL)2.50 M=26.25 mL$

Hence volume of $HCl$ required is $26.25 mL$ .

(d)

Interpretation Introduction

Interpretation:

Molarity of $HCl$ solution if its $22.20 mL$ reacts with $15.60 mL$ of $0.250 M KMnO4$ has to be determined.

Concept Introduction:

Refer to part (c).

(d)

Expert Solution

Explanation of Solution

Rearrange equation (5) for $M2$ .

$M2=M1V1V2$ (7)

Substitute $0.250 M$ for $M1$ , $15.60 mL$ for $V1$ and $22.20 mL$ for $V2$ in equation (7) for molarity of $HCl$ .

$Molarity of HCl=(0.250 M)(15.60 mL)22.20 mL=0.176 M$

Hence molarity of $HCl$ required is $0.176 M$ .

(e)

Interpretation Introduction

Interpretation:

Volume $(L)$ of $Cl2$ gas at STP produced by reaction of $125 mL$ of $2.5 M HCl$ has to be determined.

Concept Introduction:

Expression for ideal gas equation is as follows:

$PV=nRT$ (8)

Here,

$P$ is pressure of gas.

$V$ is volume of gas.

$n$ is moles of gas.

$R$ is universal gas constant.

$T$ is absolute temperature of gas.

(e)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

Expression for molarity of $HCl$ is as follows:

$Molarity of HCl=Moles of HClVolume(L) of HCl solution$ (9)

Rearrange equation (9) for moles of $HCl$ .

$Moles of HCl=[(Molarity of HCl)(Volume of HCl solution)]$ (10)

Substitute $2.5 M$ for molarity and $125 mL$ for volume of $HCl$ solution in equation (10).

$Moles of HCl=(2.5 M)(125 mL)(10−3 L1 mL)=0.3125 mol$

Since 16 moles of $HCl$ produce 5 moles of $Cl2$ , amount of $Cl2$ produced by 0.3125 moles of $HCl$ is calculated as follows:

$Amount of Cl2=(5 mol Cl216 mol HCl)(0.3125 mol HCl)=0.0976 mol$

Rearrange equation (8) for $V$ .

$V=nRTP$ (11)

Substitute $273 K$ for $T$ , $0.0976 mol$ for $n$ , $0.082057 L⋅atm⋅K−1⋅mol−1$ for $R$ and $1 atm$ for $P$ in equation (11) for volume of $Cl2$ .

$Volume of Cl2=(0.0976 mol)(0.082057 L⋅atm⋅K−1⋅mol−1)(273 K)1 atm=2.186 L$

Hence, volume of $Cl2$ is $2.186 L$ .

(f)

Interpretation Introduction

Interpretation:

Volume $(L)$ of $Cl2$ gas at STP produced by reaction of $15.0 mL$ of $0.750 M HCl$ and $12.0 mL$ of $0.550 M KMnO4$ has to be determined.

Concept Introduction:

Refer to part (e).

(f)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

Expression for molarity of $HCl$ is as follows:

$Molarity of HCl=Moles of HClVolume(L) of HCl solution$ (9)

Rearrange equation (9) for moles of $HCl$ .

$Moles of HCl=[(Molarity of HCl)(Volume of HCl solution)]$ (10)

Substitute $0.750 M$ for molarity and $15.0 mL$ for volume of $HCl$ solution in equation (10).

$Moles of HCl=(0.750 M)(15.0 mL)(10−3 L1 mL)=0.01125 mol$

Expression for molarity of $KMnO4$ is as follows:

$Molarity of KMnO4=Moles of KMnO4Volume(L) of KMnO4$ (3)

Rearrange equation (3) for moles of $KMnO4$ .

$Moles of KMnO4=[(Molarity of KMnO4)(Volume of KMnO4)]$ (12)

Substitute $12.0 mL$ for molarity and $0.550 M$ for molarity of $KMnO4$ in equation (12).

$Moles of KMnO4=(0.550 M)(12.0 mL)(10−3 L1 mL)=0.0066 mol$

Since 16 moles of $HCl$ form 5 moles of $Cl2$ , amount of $Cl2$ produced by 0.01125 moles of $HCl$ is calculated as follows:

$Amount of Cl2=(5 mol Cl216 mol HCl)(0.01125 mol HCl)=0.003516 mol$

Since 2 moles of $KMnO4$ form 5 moles of $Cl2$ , amount of $Cl2$ produced by 0.0066 moles of $KMnO4$ is calculated as follows:

$Amount of Cl2=(5 mol Cl22 mol KMnO4)(0.0066 mol KMnO4)=0.0165 mol$

Since $HCl$ produces less amount of $Cl2$ than $KMnO4$ , $HCl$ acts as limiting reactant and amount of $Cl2$ will be produced in accordance with its amount only.

Rearrange equation (8) for $V$ .

$V=nRTP$ (11)

Substitute $273 K$ for $T$ , $0.003516 mol$ for $n$ , $0.082057 L⋅atm⋅K−1⋅mol−1$ for $R$ and $1 atm$ for $P$ in equation (11) for volume of $Cl2$ .

$Volume of Cl2=(0.003516 mol)(0.082057 L⋅atm⋅K−1⋅mol−1)(273 K)1 atm=0.079 L$

Hence, volume of $Cl2$ is $0.079 L$ .

Answer 6

Chapter 14, Problem 37PE

(a)

Interpretation Introduction

Interpretation:

Moles of $H2O$ produced from $15.0 mL$ of $0.250 M HCl$ have to be determined.

Concept Introduction:

Stoichiometry describes quantitative relationships between reactants and products in any chemical reactions. In these problems, amount of one species is determined with help of known amount of another species.

(a)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

Expression for molarity of $HCl$ is as follows:

$Molarity of HCl=Moles of HClVolume(L) of HCl$ (1)

Rearrange equation (1) for moles of $HCl$ .

$Moles of HCl=(Molarity of HCl)(Volume of HCl)$ (2)

Substitute $15.0 mL$ for volume and $0.250 M$ for molarity of $HCl$ in equation (2).

$Moles of HCl=(0.250 M)(15.0 mL)(10−3 L1 mL)=0.00375 mol$

According to balanced chemical equation, two moles of $KMnO4$ react with sixteen moles of $HCl$ and produce two moles of $MnCl2$ , eight moles of $H2O$ and two moles of $KCl$ . Moles of $H2O$ produced from 0.00375 moles of $HCl$ is calculated as follows:

$Moles of H2O=(8 mol H2O16 mol HCl)(0.00375 mol HCl)=0.001875 mol$

Hence, amount of $H2O$ is $0.001875 mol$ .

Answer 7

Chapter 14, Problem 37PE

(a)

Interpretation Introduction

Interpretation:

Moles of $H2O$ produced from $15.0 mL$ of $0.250 M HCl$ have to be determined.

Concept Introduction:

Stoichiometry describes quantitative relationships between reactants and products in any chemical reactions. In these problems, amount of one species is determined with help of known amount of another species.

Answer 8

(a)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

Expression for molarity of $HCl$ is as follows:

$Molarity of HCl=Moles of HClVolume(L) of HCl$ (1)

Rearrange equation (1) for moles of $HCl$ .

$Moles of HCl=(Molarity of HCl)(Volume of HCl)$ (2)

Substitute $15.0 mL$ for volume and $0.250 M$ for molarity of $HCl$ in equation (2).

$Moles of HCl=(0.250 M)(15.0 mL)(10−3 L1 mL)=0.00375 mol$

According to balanced chemical equation, two moles of $KMnO4$ react with sixteen moles of $HCl$ and produce two moles of $MnCl2$ , eight moles of $H2O$ and two moles of $KCl$ . Moles of $H2O$ produced from 0.00375 moles of $HCl$ is calculated as follows:

$Moles of H2O=(8 mol H2O16 mol HCl)(0.00375 mol HCl)=0.001875 mol$

Hence, amount of $H2O$ is $0.001875 mol$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

Volume of $0.150 M KMnO4$ produced from $1.85 mol MnCl2$ have to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

According to balanced chemical equation, two moles of $KMnO4$ react with sixteen moles of $HCl$ and produce two moles of $MnCl2$ , eight moles of $H2O$ and two moles of $KCl$ . Amount of $KMnO4$ produced by 1.85 moles of $MnCl2$ is calculated as follows:

$Moles of KMnO4=(2 mol KMnO42 mol MnCl2)(1.85 mol MnCl2)=1.85 mol$

Expression for molarity of $KMnO4$ is as follows:

$Molarity of KMnO4=Moles of KMnO4Volume(L) of KMnO4$ (3)

Rearrange equation (3) for volume of $KMnO4$ .

$Volume of KMnO4=Moles of KMnO4Molarity of KMnO4$ (4)

Substitute $1.85 mol$ for moles and $0.150 M$ for molarity of $KMnO4$ in equation (4).

$Volume of KMnO4=1.85 mol0.150 M=12.33 L$

Hence, volume of $KMnO4$ is $12.33 L$ .

Answer 13

(b)

Interpretation Introduction

Interpretation:

Volume of $0.150 M KMnO4$ produced from $1.85 mol MnCl2$ have to be determined.

Concept Introduction:

Refer to part (a).

Answer 14

(b)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

According to balanced chemical equation, two moles of $KMnO4$ react with sixteen moles of $HCl$ and produce two moles of $MnCl2$ , eight moles of $H2O$ and two moles of $KCl$ . Amount of $KMnO4$ produced by 1.85 moles of $MnCl2$ is calculated as follows:

$Moles of KMnO4=(2 mol KMnO42 mol MnCl2)(1.85 mol MnCl2)=1.85 mol$

Expression for molarity of $KMnO4$ is as follows:

$Molarity of KMnO4=Moles of KMnO4Volume(L) of KMnO4$ (3)

Rearrange equation (3) for volume of $KMnO4$ .

$Volume of KMnO4=Moles of KMnO4Molarity of KMnO4$ (4)

Substitute $1.85 mol$ for moles and $0.150 M$ for molarity of $KMnO4$ in equation (4).

$Volume of KMnO4=1.85 mol0.150 M=12.33 L$

Hence, volume of $KMnO4$ is $12.33 L$ .

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

Volume of $2.50 M HCl$ required to produce $125 mL$ of $0.525 M KCl$ have to be determined.

Concept Introduction:

Expression for molarity equation is as follows:

$M1V1=M2V2$ (5)

Here,

$M1$ is molarity of one solution.

$V1$ is volume of one solution.

$M2$ is molarity of another solution.

$V2$ is volume of another solution.

(c)

Expert Solution

Explanation of Solution

Rearrange equation (5) for $V2$ .

$V2=M1V1M2$ (6)

Substitute $0.525 M$ for $M1$ , $125 mL$ for $V1$ and $2.50 M$ for $M2$ in equation (6) for volume of $HCl$ .

$Volume of HCl=(0.525 M)(125 mL)2.50 M=26.25 mL$

Hence volume of $HCl$ required is $26.25 mL$ .

Answer 19

(c)

Interpretation Introduction

Interpretation:

Volume of $2.50 M HCl$ required to produce $125 mL$ of $0.525 M KCl$ have to be determined.

Concept Introduction:

Expression for molarity equation is as follows:

$M1V1=M2V2$ (5)

Here,

$M1$ is molarity of one solution.

$V1$ is volume of one solution.

$M2$ is molarity of another solution.

$V2$ is volume of another solution.

Answer 20

(c)

Expert Solution

Explanation of Solution

Rearrange equation (5) for $V2$ .

$V2=M1V1M2$ (6)

Substitute $0.525 M$ for $M1$ , $125 mL$ for $V1$ and $2.50 M$ for $M2$ in equation (6) for volume of $HCl$ .

$Volume of HCl=(0.525 M)(125 mL)2.50 M=26.25 mL$

Hence volume of $HCl$ required is $26.25 mL$ .

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

(d)

Interpretation Introduction

Interpretation:

Molarity of $HCl$ solution if its $22.20 mL$ reacts with $15.60 mL$ of $0.250 M KMnO4$ has to be determined.

Concept Introduction:

Refer to part (c).

(d)

Expert Solution

Explanation of Solution

Rearrange equation (5) for $M2$ .

$M2=M1V1V2$ (7)

Substitute $0.250 M$ for $M1$ , $15.60 mL$ for $V1$ and $22.20 mL$ for $V2$ in equation (7) for molarity of $HCl$ .

$Molarity of HCl=(0.250 M)(15.60 mL)22.20 mL=0.176 M$

Hence molarity of $HCl$ required is $0.176 M$ .

Answer 25

(d)

Interpretation Introduction

Interpretation:

Molarity of $HCl$ solution if its $22.20 mL$ reacts with $15.60 mL$ of $0.250 M KMnO4$ has to be determined.

Concept Introduction:

Refer to part (c).

Answer 26

(d)

Expert Solution

Explanation of Solution

Rearrange equation (5) for $M2$ .

$M2=M1V1V2$ (7)

Substitute $0.250 M$ for $M1$ , $15.60 mL$ for $V1$ and $22.20 mL$ for $V2$ in equation (7) for molarity of $HCl$ .

$Molarity of HCl=(0.250 M)(15.60 mL)22.20 mL=0.176 M$

Hence molarity of $HCl$ required is $0.176 M$ .

Answer 27

(d)

Expert Solution

Answer 28

(d)

Expert Solution

Answer 29

Expert Solution

Answer 30

(e)

Interpretation Introduction

Interpretation:

Volume $(L)$ of $Cl2$ gas at STP produced by reaction of $125 mL$ of $2.5 M HCl$ has to be determined.

Concept Introduction:

Expression for ideal gas equation is as follows:

$PV=nRT$ (8)

Here,

$P$ is pressure of gas.

$V$ is volume of gas.

$n$ is moles of gas.

$R$ is universal gas constant.

$T$ is absolute temperature of gas.

(e)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

Expression for molarity of $HCl$ is as follows:

$Molarity of HCl=Moles of HClVolume(L) of HCl solution$ (9)

Rearrange equation (9) for moles of $HCl$ .

$Moles of HCl=[(Molarity of HCl)(Volume of HCl solution)]$ (10)

Substitute $2.5 M$ for molarity and $125 mL$ for volume of $HCl$ solution in equation (10).

$Moles of HCl=(2.5 M)(125 mL)(10−3 L1 mL)=0.3125 mol$

Since 16 moles of $HCl$ produce 5 moles of $Cl2$ , amount of $Cl2$ produced by 0.3125 moles of $HCl$ is calculated as follows:

$Amount of Cl2=(5 mol Cl216 mol HCl)(0.3125 mol HCl)=0.0976 mol$

Rearrange equation (8) for $V$ .

$V=nRTP$ (11)

Substitute $273 K$ for $T$ , $0.0976 mol$ for $n$ , $0.082057 L⋅atm⋅K−1⋅mol−1$ for $R$ and $1 atm$ for $P$ in equation (11) for volume of $Cl2$ .

$Volume of Cl2=(0.0976 mol)(0.082057 L⋅atm⋅K−1⋅mol−1)(273 K)1 atm=2.186 L$

Hence, volume of $Cl2$ is $2.186 L$ .

Answer 31

(e)

Interpretation Introduction

Interpretation:

Volume $(L)$ of $Cl2$ gas at STP produced by reaction of $125 mL$ of $2.5 M HCl$ has to be determined.

Concept Introduction:

Expression for ideal gas equation is as follows:

$PV=nRT$ (8)

Here,

$P$ is pressure of gas.

$V$ is volume of gas.

$n$ is moles of gas.

$R$ is universal gas constant.

$T$ is absolute temperature of gas.

Answer 32

(e)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

Expression for molarity of $HCl$ is as follows:

$Molarity of HCl=Moles of HClVolume(L) of HCl solution$ (9)

Rearrange equation (9) for moles of $HCl$ .

$Moles of HCl=[(Molarity of HCl)(Volume of HCl solution)]$ (10)

Substitute $2.5 M$ for molarity and $125 mL$ for volume of $HCl$ solution in equation (10).

$Moles of HCl=(2.5 M)(125 mL)(10−3 L1 mL)=0.3125 mol$

Since 16 moles of $HCl$ produce 5 moles of $Cl2$ , amount of $Cl2$ produced by 0.3125 moles of $HCl$ is calculated as follows:

$Amount of Cl2=(5 mol Cl216 mol HCl)(0.3125 mol HCl)=0.0976 mol$

Rearrange equation (8) for $V$ .

$V=nRTP$ (11)

Substitute $273 K$ for $T$ , $0.0976 mol$ for $n$ , $0.082057 L⋅atm⋅K−1⋅mol−1$ for $R$ and $1 atm$ for $P$ in equation (11) for volume of $Cl2$ .

$Volume of Cl2=(0.0976 mol)(0.082057 L⋅atm⋅K−1⋅mol−1)(273 K)1 atm=2.186 L$

Hence, volume of $Cl2$ is $2.186 L$ .

Answer 33

(e)

Expert Solution

Answer 34

(e)

Expert Solution

Answer 35

Expert Solution

Answer 36

(f)

Interpretation Introduction

Interpretation:

Volume $(L)$ of $Cl2$ gas at STP produced by reaction of $15.0 mL$ of $0.750 M HCl$ and $12.0 mL$ of $0.550 M KMnO4$ has to be determined.

Concept Introduction:

Refer to part (e).

(f)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

Expression for molarity of $HCl$ is as follows:

$Molarity of HCl=Moles of HClVolume(L) of HCl solution$ (9)

Rearrange equation (9) for moles of $HCl$ .

$Moles of HCl=[(Molarity of HCl)(Volume of HCl solution)]$ (10)

Substitute $0.750 M$ for molarity and $15.0 mL$ for volume of $HCl$ solution in equation (10).

$Moles of HCl=(0.750 M)(15.0 mL)(10−3 L1 mL)=0.01125 mol$

Expression for molarity of $KMnO4$ is as follows:

$Molarity of KMnO4=Moles of KMnO4Volume(L) of KMnO4$ (3)

Rearrange equation (3) for moles of $KMnO4$ .

$Moles of KMnO4=[(Molarity of KMnO4)(Volume of KMnO4)]$ (12)

Substitute $12.0 mL$ for molarity and $0.550 M$ for molarity of $KMnO4$ in equation (12).

$Moles of KMnO4=(0.550 M)(12.0 mL)(10−3 L1 mL)=0.0066 mol$

Since 16 moles of $HCl$ form 5 moles of $Cl2$ , amount of $Cl2$ produced by 0.01125 moles of $HCl$ is calculated as follows:

$Amount of Cl2=(5 mol Cl216 mol HCl)(0.01125 mol HCl)=0.003516 mol$

Since 2 moles of $KMnO4$ form 5 moles of $Cl2$ , amount of $Cl2$ produced by 0.0066 moles of $KMnO4$ is calculated as follows:

$Amount of Cl2=(5 mol Cl22 mol KMnO4)(0.0066 mol KMnO4)=0.0165 mol$

Since $HCl$ produces less amount of $Cl2$ than $KMnO4$ , $HCl$ acts as limiting reactant and amount of $Cl2$ will be produced in accordance with its amount only.

Rearrange equation (8) for $V$ .

$V=nRTP$ (11)

Substitute $273 K$ for $T$ , $0.003516 mol$ for $n$ , $0.082057 L⋅atm⋅K−1⋅mol−1$ for $R$ and $1 atm$ for $P$ in equation (11) for volume of $Cl2$ .

$Volume of Cl2=(0.003516 mol)(0.082057 L⋅atm⋅K−1⋅mol−1)(273 K)1 atm=0.079 L$

Hence, volume of $Cl2$ is $0.079 L$ .

Answer 37

(f)

Interpretation Introduction

Interpretation:

Volume $(L)$ of $Cl2$ gas at STP produced by reaction of $15.0 mL$ of $0.750 M HCl$ and $12.0 mL$ of $0.550 M KMnO4$ has to be determined.

Concept Introduction:

Refer to part (e).

Answer 38

(f)

Expert Solution

Explanation of Solution

Given reaction occurs as follows:

$2KMnO4(aq)+16HCl(aq)→2MnCl2(aq)+5Cl2(g)+8H2O(l)+2KCl(aq)$

Expression for molarity of $HCl$ is as follows:

$Molarity of HCl=Moles of HClVolume(L) of HCl solution$ (9)

Rearrange equation (9) for moles of $HCl$ .

$Moles of HCl=[(Molarity of HCl)(Volume of HCl solution)]$ (10)

Substitute $0.750 M$ for molarity and $15.0 mL$ for volume of $HCl$ solution in equation (10).

$Moles of HCl=(0.750 M)(15.0 mL)(10−3 L1 mL)=0.01125 mol$

Expression for molarity of $KMnO4$ is as follows:

$Molarity of KMnO4=Moles of KMnO4Volume(L) of KMnO4$ (3)

Rearrange equation (3) for moles of $KMnO4$ .

$Moles of KMnO4=[(Molarity of KMnO4)(Volume of KMnO4)]$ (12)

Substitute $12.0 mL$ for molarity and $0.550 M$ for molarity of $KMnO4$ in equation (12).

$Moles of KMnO4=(0.550 M)(12.0 mL)(10−3 L1 mL)=0.0066 mol$

Since 16 moles of $HCl$ form 5 moles of $Cl2$ , amount of $Cl2$ produced by 0.01125 moles of $HCl$ is calculated as follows:

$Amount of Cl2=(5 mol Cl216 mol HCl)(0.01125 mol HCl)=0.003516 mol$

Since 2 moles of $KMnO4$ form 5 moles of $Cl2$ , amount of $Cl2$ produced by 0.0066 moles of $KMnO4$ is calculated as follows:

$Amount of Cl2=(5 mol Cl22 mol KMnO4)(0.0066 mol KMnO4)=0.0165 mol$

Since $HCl$ produces less amount of $Cl2$ than $KMnO4$ , $HCl$ acts as limiting reactant and amount of $Cl2$ will be produced in accordance with its amount only.