
Concept explainers
(a)
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
(a)

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:
Volume of 0.150 M KMnO4 produced from 1.85 mol MnCl2 have to be determined.
Concept Introduction:
Refer to part (a).
(b)

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:
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)

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:
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)

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:
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)

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:
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)

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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Chapter 14 Solutions
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