The pair of combination that produces a greater mass of the solid product is to be determined. Concept introduction: In a precipitation reaction, two ionic compounds in a solution react to form an insoluble compound, which is called the precipitate. The formation of the precipitate can be predicted by using solubility rules. The mass of the precipitate can be determined from the stoichiometry of the reaction.

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Answer 1

Question

Chapter 4, Problem 91QP

Interpretation Introduction

Interpretation:

The pair of combination that produces a greater mass of the solid product is to be determined.

Concept introduction:

In a precipitation reaction, two ionic compounds in a solution react to form an insoluble compound, which is called the precipitate.

The formation of the precipitate can be predicted by using solubility rules.

The mass of the precipitate can be determined from the stoichiometry of the reaction.

Expert Solution & Answer

Answer to Problem 91QP

Solution:

(a) Combination 1

(b) Combination 2

(c) Combination 2

Explanation of Solution

a)

Combination 1: $105.5 mL 1.508 M Pb(NO3)2and 250.0 mL 1.2075 M KCl$

Combination 2: $138.5 mL 1.469 M Pb(NO3)2and 100.0 mL 2.115 M KCl$

In Combination 1:

The precipitation reaction between lead nitrate and potassium chloride is as follows:

$2KCl(aq)+Pb(NO3)2(aq)→PbCl2(s)+2KNO3(aq)$

From the above balanced equation, one mole of $Pb(NO3)2$ reacts with two moles of KCl.

First, calculate the number of moles of $Pb(NO3)2$ using the given volume and concentration:

$Moles of Pb(NO3)2= 1.508 M×0.1055 L=0.159094 mol$

Now, calculate the number of moles of KCl.

$Moles of KCl = 1.2075 M×0.250 L=0.301875 mol$

Hence, KCl is a limiting reagent that determines the amount of $PbCl2$ formed.

$Moles of PbCl2 = mol KCl×1 mol PbCl22 mol KCl=0.301875 mol×1 mol PbCl22 mol KCl=0.150938 mol$

Convert moles to grams by multiplying the moles with the molar mass of $PbCl2$

$Mass of PbCl2 = 0.150938 mol×278.1 g/mol PbCl2 = 41.98 g$

Hence, the mass of the $PbCl2$ formed is $41.98 g$ .

In Combination 2,

Calculate the number of moles of $Pb(NO3)2$ using the given volume and concentration

$Moles of Pb(NO3)2 = 1.469 M×0.1385 L=0.203457 mol$

Now, calculate the number of moles of KCl.

$Moles of KCl = 2.115 M×0.100 L=0.2115 mol$

KCl is a limiting reagent that determines the amount of $PbCl2$ formed.

$Moles of PbCl2 = mol KCl×1 mol PbCl22 mol KCl=0.2115 mol×1 mol PbCl22 mol KCl=0.10575 mol$

Now convert the number of moles to grams by multiplying the number of moles with the molar mass of PbCl₂

$g PbCl2= 0.10575 mol×278.1 g/mol PbCl2 = 29.41 g$

Hence, the mass of the PbCl₂ formed is $29.41 g$ .

Therefore, combination 1 will produce a greater mass of the product.

b)

Combination 1: $32.25 mL 0.9475M Na3PO4 and 92.75 mL 0.7750M Ca(NO3)2$

Combination 2: $52.50 mL 0.6810M Na3PO4 and 39.50 mL 1.555M Ca(NO3)2$

Combination 1:

The precipitation reaction between sodium phosphate and calcium nitrate is as follows:

$2Na3PO4(aq)+3Ca(NO3)2(aq)→Ca3(PO4)2(s)+6NaNO3(aq)$

From the above balanced equation, two moles of $Na3PO4$ react with three moles of $Ca(NO3)2$ .

First, calculate the number of moles of Na₃PO₄ using the given volume and concentration:

$Moles of Na3PO4= 0.9475 M×0.03225 L=0.030557 mol$

Now, calculate the number of moles of $Ca(NO3)2$ :

$Moles of Ca(NO3)2= 0.7750 M×0.09275 L=0.071881 mol$

Here, Na₃PO₄ is a limiting reagent that determines the amount of product formed.

$Moles of Ca3(PO4)2 = mol Na3PO4×1 mol Ca3(PO4)22 mol Na3PO4 =0.030557 mol×1 mol Ca3(PO4)22 mol Na3PO4 =0.015279 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of $Ca3(PO4)2$ :

$Mass of Ca3(PO4)2 = 0.015279 mol × 310.18 g/mol Ca3(PO4)2= 4.74 g$

Hence, the mass of the $Ca3(PO4)2$ formed is $4.74 g$ .

Combination 2:

Calculate the number of moles of $Na3PO4$ using the given volume and concentration:

$Moles of Na3PO4 = 0.6810 M × 0.05250 L=0.0357525 mol$

Now, calculate the number of moles of $Ca(NO3)2$ :

$Moles of Ca(NO3)2= 0.7750 M×0.09275 L=0.0614225 mol$

Here, $Na3PO4$ is a limiting reagent that determines the amount of product formed.

$Moles of Ca3(PO4)2 = mol Na3PO4×1 mol Ca3(PO4)22 mol Na3PO4 =0.0357525 mol×1 mol Ca3(PO4)22 mol Na3PO4 =0.01787625 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of $Ca3(PO4)2$ :

$g Ca3(PO4)2= 0.01787625 mol × 310.18 g/mol Ca3(PO4)2= 5.54 g$

Hence, the mass of the $Ca3(PO4)2$ formed is $5.54 g$ .

Therefore, combination 2 produces a greater mass of the product.

c)

Combination 1: $29.75 mL 1.575 M AgNO3 and 25.00 mL 2.010 M BaCl2$

Combination 2: $52.80 mL 2.010 M AgNO3 and 73.50 mL 0.7500 M BaCl2$

Combination 1:

The precipitation reaction between silver nitrate and barium chloride is as follows:

$BaCl2(aq)+2AgNO3(aq)→2AgCl(s)+Ba(NO3)2(aq)$

From the above balanced equation, one mole of BaCl2 reacts with two moles of $AgNO3$ .

First, calculate the number of moles of $BaCl2$ using the given volume and concentration:

$Moles of BaCl2 = 2.010 M×0.0250 L=0.05025 mol$

Now, calculate the number of moles of $AgNO3$ .

$Moles of AgNO3 = 1.575 M×0.02975 L=0.0468563 mol$

Hence, $AgNO3$ is a limiting reagent that determines the amount of AgCl formed.

$Moles of AgCl = mol AgNO3×2 mol AgCl2 mol AgNO3=0.0468563 mol AgNO3×2 mol AgCl2 mol AgNO3=0.0468563 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of AgCl:

$g AgCl = 0.0468563 mol × 143.32 g/mol AgCl = 6.72 g$

Hence, the mass of the AgCl formed is $6.72 g$ .

Combination 2:

Calculate the number of moles of BaCl₂ using the given volume and concentration:

$Moles of BaCl2 = 0.7500 M×0.07350 L=0.055125 mol$

Now, calculate the number of moles of AgNO₃.

$Moles of AgNO3 = 2.010 M×0.05280 L=0.106128 mol$

Hence, $AgNO3$ is a limiting reagent that determines the amount of AgCl formed.

$Moles of AgCl = mol AgNO3×2 mol AgCl2 mol AgNO3=0.106128 mol AgNO3×2 mol AgCl2 mol AgNO3=0.106128 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of AgCl:

$g AgCl = 0.106128 mol×143.32 g/mol AgCl = 15.21 g$

Hence, the mass of the AgCl formed is $15.21 g$ .

Therefore, combination 2 produces a greater mass of the product.

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Answer 2

Question

Chapter 4, Problem 91QP

Interpretation Introduction

Interpretation:

The pair of combination that produces a greater mass of the solid product is to be determined.

Concept introduction:

In a precipitation reaction, two ionic compounds in a solution react to form an insoluble compound, which is called the precipitate.

The formation of the precipitate can be predicted by using solubility rules.

The mass of the precipitate can be determined from the stoichiometry of the reaction.

Expert Solution & Answer

Answer to Problem 91QP

Solution:

(a) Combination 1

(b) Combination 2

(c) Combination 2

Explanation of Solution

a)

Combination 1: $105.5 mL 1.508 M Pb(NO3)2and 250.0 mL 1.2075 M KCl$

Combination 2: $138.5 mL 1.469 M Pb(NO3)2and 100.0 mL 2.115 M KCl$

In Combination 1:

The precipitation reaction between lead nitrate and potassium chloride is as follows:

$2KCl(aq)+Pb(NO3)2(aq)→PbCl2(s)+2KNO3(aq)$

From the above balanced equation, one mole of $Pb(NO3)2$ reacts with two moles of KCl.

First, calculate the number of moles of $Pb(NO3)2$ using the given volume and concentration:

$Moles of Pb(NO3)2= 1.508 M×0.1055 L=0.159094 mol$

Now, calculate the number of moles of KCl.

$Moles of KCl = 1.2075 M×0.250 L=0.301875 mol$

Hence, KCl is a limiting reagent that determines the amount of $PbCl2$ formed.

$Moles of PbCl2 = mol KCl×1 mol PbCl22 mol KCl=0.301875 mol×1 mol PbCl22 mol KCl=0.150938 mol$

Convert moles to grams by multiplying the moles with the molar mass of $PbCl2$

$Mass of PbCl2 = 0.150938 mol×278.1 g/mol PbCl2 = 41.98 g$

Hence, the mass of the $PbCl2$ formed is $41.98 g$ .

In Combination 2,

Calculate the number of moles of $Pb(NO3)2$ using the given volume and concentration

$Moles of Pb(NO3)2 = 1.469 M×0.1385 L=0.203457 mol$

Now, calculate the number of moles of KCl.

$Moles of KCl = 2.115 M×0.100 L=0.2115 mol$

KCl is a limiting reagent that determines the amount of $PbCl2$ formed.

$Moles of PbCl2 = mol KCl×1 mol PbCl22 mol KCl=0.2115 mol×1 mol PbCl22 mol KCl=0.10575 mol$

Now convert the number of moles to grams by multiplying the number of moles with the molar mass of PbCl₂

$g PbCl2= 0.10575 mol×278.1 g/mol PbCl2 = 29.41 g$

Hence, the mass of the PbCl₂ formed is $29.41 g$ .

Therefore, combination 1 will produce a greater mass of the product.

b)

Combination 1: $32.25 mL 0.9475M Na3PO4 and 92.75 mL 0.7750M Ca(NO3)2$

Combination 2: $52.50 mL 0.6810M Na3PO4 and 39.50 mL 1.555M Ca(NO3)2$

Combination 1:

The precipitation reaction between sodium phosphate and calcium nitrate is as follows:

$2Na3PO4(aq)+3Ca(NO3)2(aq)→Ca3(PO4)2(s)+6NaNO3(aq)$

From the above balanced equation, two moles of $Na3PO4$ react with three moles of $Ca(NO3)2$ .

First, calculate the number of moles of Na₃PO₄ using the given volume and concentration:

$Moles of Na3PO4= 0.9475 M×0.03225 L=0.030557 mol$

Now, calculate the number of moles of $Ca(NO3)2$ :

$Moles of Ca(NO3)2= 0.7750 M×0.09275 L=0.071881 mol$

Here, Na₃PO₄ is a limiting reagent that determines the amount of product formed.

$Moles of Ca3(PO4)2 = mol Na3PO4×1 mol Ca3(PO4)22 mol Na3PO4 =0.030557 mol×1 mol Ca3(PO4)22 mol Na3PO4 =0.015279 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of $Ca3(PO4)2$ :

$Mass of Ca3(PO4)2 = 0.015279 mol × 310.18 g/mol Ca3(PO4)2= 4.74 g$

Hence, the mass of the $Ca3(PO4)2$ formed is $4.74 g$ .

Combination 2:

Calculate the number of moles of $Na3PO4$ using the given volume and concentration:

$Moles of Na3PO4 = 0.6810 M × 0.05250 L=0.0357525 mol$

Now, calculate the number of moles of $Ca(NO3)2$ :

$Moles of Ca(NO3)2= 0.7750 M×0.09275 L=0.0614225 mol$

Here, $Na3PO4$ is a limiting reagent that determines the amount of product formed.

$Moles of Ca3(PO4)2 = mol Na3PO4×1 mol Ca3(PO4)22 mol Na3PO4 =0.0357525 mol×1 mol Ca3(PO4)22 mol Na3PO4 =0.01787625 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of $Ca3(PO4)2$ :

$g Ca3(PO4)2= 0.01787625 mol × 310.18 g/mol Ca3(PO4)2= 5.54 g$

Hence, the mass of the $Ca3(PO4)2$ formed is $5.54 g$ .

Therefore, combination 2 produces a greater mass of the product.

c)

Combination 1: $29.75 mL 1.575 M AgNO3 and 25.00 mL 2.010 M BaCl2$

Combination 2: $52.80 mL 2.010 M AgNO3 and 73.50 mL 0.7500 M BaCl2$

Combination 1:

The precipitation reaction between silver nitrate and barium chloride is as follows:

$BaCl2(aq)+2AgNO3(aq)→2AgCl(s)+Ba(NO3)2(aq)$

From the above balanced equation, one mole of BaCl2 reacts with two moles of $AgNO3$ .

First, calculate the number of moles of $BaCl2$ using the given volume and concentration:

$Moles of BaCl2 = 2.010 M×0.0250 L=0.05025 mol$

Now, calculate the number of moles of $AgNO3$ .

$Moles of AgNO3 = 1.575 M×0.02975 L=0.0468563 mol$

Hence, $AgNO3$ is a limiting reagent that determines the amount of AgCl formed.

$Moles of AgCl = mol AgNO3×2 mol AgCl2 mol AgNO3=0.0468563 mol AgNO3×2 mol AgCl2 mol AgNO3=0.0468563 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of AgCl:

$g AgCl = 0.0468563 mol × 143.32 g/mol AgCl = 6.72 g$

Hence, the mass of the AgCl formed is $6.72 g$ .

Combination 2:

Calculate the number of moles of BaCl₂ using the given volume and concentration:

$Moles of BaCl2 = 0.7500 M×0.07350 L=0.055125 mol$

Now, calculate the number of moles of AgNO₃.

$Moles of AgNO3 = 2.010 M×0.05280 L=0.106128 mol$

Hence, $AgNO3$ is a limiting reagent that determines the amount of AgCl formed.

$Moles of AgCl = mol AgNO3×2 mol AgCl2 mol AgNO3=0.106128 mol AgNO3×2 mol AgCl2 mol AgNO3=0.106128 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of AgCl:

$g AgCl = 0.106128 mol×143.32 g/mol AgCl = 15.21 g$

Hence, the mass of the AgCl formed is $15.21 g$ .

Therefore, combination 2 produces a greater mass of the product.

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Answer 3

Question

Chapter 4, Problem 91QP

Interpretation Introduction

Interpretation:

The pair of combination that produces a greater mass of the solid product is to be determined.

Concept introduction:

In a precipitation reaction, two ionic compounds in a solution react to form an insoluble compound, which is called the precipitate.

The formation of the precipitate can be predicted by using solubility rules.

The mass of the precipitate can be determined from the stoichiometry of the reaction.

Expert Solution & Answer

Answer to Problem 91QP

Solution:

(a) Combination 1

(b) Combination 2

(c) Combination 2

Explanation of Solution

a)

Combination 1: $105.5 mL 1.508 M Pb(NO3)2and 250.0 mL 1.2075 M KCl$

Combination 2: $138.5 mL 1.469 M Pb(NO3)2and 100.0 mL 2.115 M KCl$

In Combination 1:

The precipitation reaction between lead nitrate and potassium chloride is as follows:

$2KCl(aq)+Pb(NO3)2(aq)→PbCl2(s)+2KNO3(aq)$

From the above balanced equation, one mole of $Pb(NO3)2$ reacts with two moles of KCl.

First, calculate the number of moles of $Pb(NO3)2$ using the given volume and concentration:

$Moles of Pb(NO3)2= 1.508 M×0.1055 L=0.159094 mol$

Now, calculate the number of moles of KCl.

$Moles of KCl = 1.2075 M×0.250 L=0.301875 mol$

Hence, KCl is a limiting reagent that determines the amount of $PbCl2$ formed.

$Moles of PbCl2 = mol KCl×1 mol PbCl22 mol KCl=0.301875 mol×1 mol PbCl22 mol KCl=0.150938 mol$

Convert moles to grams by multiplying the moles with the molar mass of $PbCl2$

$Mass of PbCl2 = 0.150938 mol×278.1 g/mol PbCl2 = 41.98 g$

Hence, the mass of the $PbCl2$ formed is $41.98 g$ .

In Combination 2,

Calculate the number of moles of $Pb(NO3)2$ using the given volume and concentration

$Moles of Pb(NO3)2 = 1.469 M×0.1385 L=0.203457 mol$

Now, calculate the number of moles of KCl.

$Moles of KCl = 2.115 M×0.100 L=0.2115 mol$

KCl is a limiting reagent that determines the amount of $PbCl2$ formed.

$Moles of PbCl2 = mol KCl×1 mol PbCl22 mol KCl=0.2115 mol×1 mol PbCl22 mol KCl=0.10575 mol$

Now convert the number of moles to grams by multiplying the number of moles with the molar mass of PbCl₂

$g PbCl2= 0.10575 mol×278.1 g/mol PbCl2 = 29.41 g$

Hence, the mass of the PbCl₂ formed is $29.41 g$ .

Therefore, combination 1 will produce a greater mass of the product.

b)

Combination 1: $32.25 mL 0.9475M Na3PO4 and 92.75 mL 0.7750M Ca(NO3)2$

Combination 2: $52.50 mL 0.6810M Na3PO4 and 39.50 mL 1.555M Ca(NO3)2$

Combination 1:

The precipitation reaction between sodium phosphate and calcium nitrate is as follows:

$2Na3PO4(aq)+3Ca(NO3)2(aq)→Ca3(PO4)2(s)+6NaNO3(aq)$

From the above balanced equation, two moles of $Na3PO4$ react with three moles of $Ca(NO3)2$ .

First, calculate the number of moles of Na₃PO₄ using the given volume and concentration:

$Moles of Na3PO4= 0.9475 M×0.03225 L=0.030557 mol$

Now, calculate the number of moles of $Ca(NO3)2$ :

$Moles of Ca(NO3)2= 0.7750 M×0.09275 L=0.071881 mol$

Here, Na₃PO₄ is a limiting reagent that determines the amount of product formed.

$Moles of Ca3(PO4)2 = mol Na3PO4×1 mol Ca3(PO4)22 mol Na3PO4 =0.030557 mol×1 mol Ca3(PO4)22 mol Na3PO4 =0.015279 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of $Ca3(PO4)2$ :

$Mass of Ca3(PO4)2 = 0.015279 mol × 310.18 g/mol Ca3(PO4)2= 4.74 g$

Hence, the mass of the $Ca3(PO4)2$ formed is $4.74 g$ .

Combination 2:

Calculate the number of moles of $Na3PO4$ using the given volume and concentration:

$Moles of Na3PO4 = 0.6810 M × 0.05250 L=0.0357525 mol$

Now, calculate the number of moles of $Ca(NO3)2$ :

$Moles of Ca(NO3)2= 0.7750 M×0.09275 L=0.0614225 mol$

Here, $Na3PO4$ is a limiting reagent that determines the amount of product formed.

$Moles of Ca3(PO4)2 = mol Na3PO4×1 mol Ca3(PO4)22 mol Na3PO4 =0.0357525 mol×1 mol Ca3(PO4)22 mol Na3PO4 =0.01787625 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of $Ca3(PO4)2$ :

$g Ca3(PO4)2= 0.01787625 mol × 310.18 g/mol Ca3(PO4)2= 5.54 g$

Hence, the mass of the $Ca3(PO4)2$ formed is $5.54 g$ .

Therefore, combination 2 produces a greater mass of the product.

c)

Combination 1: $29.75 mL 1.575 M AgNO3 and 25.00 mL 2.010 M BaCl2$

Combination 2: $52.80 mL 2.010 M AgNO3 and 73.50 mL 0.7500 M BaCl2$

Combination 1:

The precipitation reaction between silver nitrate and barium chloride is as follows:

$BaCl2(aq)+2AgNO3(aq)→2AgCl(s)+Ba(NO3)2(aq)$

From the above balanced equation, one mole of BaCl2 reacts with two moles of $AgNO3$ .

First, calculate the number of moles of $BaCl2$ using the given volume and concentration:

$Moles of BaCl2 = 2.010 M×0.0250 L=0.05025 mol$

Now, calculate the number of moles of $AgNO3$ .

$Moles of AgNO3 = 1.575 M×0.02975 L=0.0468563 mol$

Hence, $AgNO3$ is a limiting reagent that determines the amount of AgCl formed.

$Moles of AgCl = mol AgNO3×2 mol AgCl2 mol AgNO3=0.0468563 mol AgNO3×2 mol AgCl2 mol AgNO3=0.0468563 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of AgCl:

$g AgCl = 0.0468563 mol × 143.32 g/mol AgCl = 6.72 g$

Hence, the mass of the AgCl formed is $6.72 g$ .

Combination 2:

Calculate the number of moles of BaCl₂ using the given volume and concentration:

$Moles of BaCl2 = 0.7500 M×0.07350 L=0.055125 mol$

Now, calculate the number of moles of AgNO₃.

$Moles of AgNO3 = 2.010 M×0.05280 L=0.106128 mol$

Hence, $AgNO3$ is a limiting reagent that determines the amount of AgCl formed.

$Moles of AgCl = mol AgNO3×2 mol AgCl2 mol AgNO3=0.106128 mol AgNO3×2 mol AgCl2 mol AgNO3=0.106128 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of AgCl:

$g AgCl = 0.106128 mol×143.32 g/mol AgCl = 15.21 g$

Hence, the mass of the AgCl formed is $15.21 g$ .

Therefore, combination 2 produces a greater mass of the product.

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Answer 4

Question

Chapter 4, Problem 91QP

Interpretation Introduction

Interpretation:

The pair of combination that produces a greater mass of the solid product is to be determined.

Concept introduction:

In a precipitation reaction, two ionic compounds in a solution react to form an insoluble compound, which is called the precipitate.

The formation of the precipitate can be predicted by using solubility rules.

The mass of the precipitate can be determined from the stoichiometry of the reaction.

Expert Solution & Answer

Answer to Problem 91QP

Solution:

(a) Combination 1

(b) Combination 2

(c) Combination 2

Explanation of Solution

a)

Combination 1: $105.5 mL 1.508 M Pb(NO3)2and 250.0 mL 1.2075 M KCl$

Combination 2: $138.5 mL 1.469 M Pb(NO3)2and 100.0 mL 2.115 M KCl$

In Combination 1:

The precipitation reaction between lead nitrate and potassium chloride is as follows:

$2KCl(aq)+Pb(NO3)2(aq)→PbCl2(s)+2KNO3(aq)$

From the above balanced equation, one mole of $Pb(NO3)2$ reacts with two moles of KCl.

First, calculate the number of moles of $Pb(NO3)2$ using the given volume and concentration:

$Moles of Pb(NO3)2= 1.508 M×0.1055 L=0.159094 mol$

Now, calculate the number of moles of KCl.

$Moles of KCl = 1.2075 M×0.250 L=0.301875 mol$

Hence, KCl is a limiting reagent that determines the amount of $PbCl2$ formed.

$Moles of PbCl2 = mol KCl×1 mol PbCl22 mol KCl=0.301875 mol×1 mol PbCl22 mol KCl=0.150938 mol$

Convert moles to grams by multiplying the moles with the molar mass of $PbCl2$

$Mass of PbCl2 = 0.150938 mol×278.1 g/mol PbCl2 = 41.98 g$

Hence, the mass of the $PbCl2$ formed is $41.98 g$ .

In Combination 2,

Calculate the number of moles of $Pb(NO3)2$ using the given volume and concentration

$Moles of Pb(NO3)2 = 1.469 M×0.1385 L=0.203457 mol$

Now, calculate the number of moles of KCl.

$Moles of KCl = 2.115 M×0.100 L=0.2115 mol$

KCl is a limiting reagent that determines the amount of $PbCl2$ formed.

$Moles of PbCl2 = mol KCl×1 mol PbCl22 mol KCl=0.2115 mol×1 mol PbCl22 mol KCl=0.10575 mol$

Now convert the number of moles to grams by multiplying the number of moles with the molar mass of PbCl₂

$g PbCl2= 0.10575 mol×278.1 g/mol PbCl2 = 29.41 g$

Hence, the mass of the PbCl₂ formed is $29.41 g$ .

Therefore, combination 1 will produce a greater mass of the product.

b)

Combination 1: $32.25 mL 0.9475M Na3PO4 and 92.75 mL 0.7750M Ca(NO3)2$

Combination 2: $52.50 mL 0.6810M Na3PO4 and 39.50 mL 1.555M Ca(NO3)2$

Combination 1:

The precipitation reaction between sodium phosphate and calcium nitrate is as follows:

$2Na3PO4(aq)+3Ca(NO3)2(aq)→Ca3(PO4)2(s)+6NaNO3(aq)$

From the above balanced equation, two moles of $Na3PO4$ react with three moles of $Ca(NO3)2$ .

First, calculate the number of moles of Na₃PO₄ using the given volume and concentration:

$Moles of Na3PO4= 0.9475 M×0.03225 L=0.030557 mol$

Now, calculate the number of moles of $Ca(NO3)2$ :

$Moles of Ca(NO3)2= 0.7750 M×0.09275 L=0.071881 mol$

Here, Na₃PO₄ is a limiting reagent that determines the amount of product formed.

$Moles of Ca3(PO4)2 = mol Na3PO4×1 mol Ca3(PO4)22 mol Na3PO4 =0.030557 mol×1 mol Ca3(PO4)22 mol Na3PO4 =0.015279 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of $Ca3(PO4)2$ :

$Mass of Ca3(PO4)2 = 0.015279 mol × 310.18 g/mol Ca3(PO4)2= 4.74 g$

Hence, the mass of the $Ca3(PO4)2$ formed is $4.74 g$ .

Combination 2:

Calculate the number of moles of $Na3PO4$ using the given volume and concentration:

$Moles of Na3PO4 = 0.6810 M × 0.05250 L=0.0357525 mol$

Now, calculate the number of moles of $Ca(NO3)2$ :

$Moles of Ca(NO3)2= 0.7750 M×0.09275 L=0.0614225 mol$

Here, $Na3PO4$ is a limiting reagent that determines the amount of product formed.

$Moles of Ca3(PO4)2 = mol Na3PO4×1 mol Ca3(PO4)22 mol Na3PO4 =0.0357525 mol×1 mol Ca3(PO4)22 mol Na3PO4 =0.01787625 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of $Ca3(PO4)2$ :

$g Ca3(PO4)2= 0.01787625 mol × 310.18 g/mol Ca3(PO4)2= 5.54 g$

Hence, the mass of the $Ca3(PO4)2$ formed is $5.54 g$ .

Therefore, combination 2 produces a greater mass of the product.

c)

Combination 1: $29.75 mL 1.575 M AgNO3 and 25.00 mL 2.010 M BaCl2$

Combination 2: $52.80 mL 2.010 M AgNO3 and 73.50 mL 0.7500 M BaCl2$

Combination 1:

The precipitation reaction between silver nitrate and barium chloride is as follows:

$BaCl2(aq)+2AgNO3(aq)→2AgCl(s)+Ba(NO3)2(aq)$

From the above balanced equation, one mole of BaCl2 reacts with two moles of $AgNO3$ .

First, calculate the number of moles of $BaCl2$ using the given volume and concentration:

$Moles of BaCl2 = 2.010 M×0.0250 L=0.05025 mol$

Now, calculate the number of moles of $AgNO3$ .

$Moles of AgNO3 = 1.575 M×0.02975 L=0.0468563 mol$

Hence, $AgNO3$ is a limiting reagent that determines the amount of AgCl formed.

$Moles of AgCl = mol AgNO3×2 mol AgCl2 mol AgNO3=0.0468563 mol AgNO3×2 mol AgCl2 mol AgNO3=0.0468563 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of AgCl:

$g AgCl = 0.0468563 mol × 143.32 g/mol AgCl = 6.72 g$

Hence, the mass of the AgCl formed is $6.72 g$ .

Combination 2:

Calculate the number of moles of BaCl₂ using the given volume and concentration:

$Moles of BaCl2 = 0.7500 M×0.07350 L=0.055125 mol$

Now, calculate the number of moles of AgNO₃.

$Moles of AgNO3 = 2.010 M×0.05280 L=0.106128 mol$

Hence, $AgNO3$ is a limiting reagent that determines the amount of AgCl formed.

$Moles of AgCl = mol AgNO3×2 mol AgCl2 mol AgNO3=0.106128 mol AgNO3×2 mol AgCl2 mol AgNO3=0.106128 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of AgCl:

$g AgCl = 0.106128 mol×143.32 g/mol AgCl = 15.21 g$

Hence, the mass of the AgCl formed is $15.21 g$ .

Therefore, combination 2 produces a greater mass of the product.

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Answer 5

Chapter 4, Problem 91QP

Interpretation Introduction

Interpretation:

The pair of combination that produces a greater mass of the solid product is to be determined.

Concept introduction:

In a precipitation reaction, two ionic compounds in a solution react to form an insoluble compound, which is called the precipitate.

The formation of the precipitate can be predicted by using solubility rules.

The mass of the precipitate can be determined from the stoichiometry of the reaction.

Expert Solution & Answer

Answer to Problem 91QP

Solution:

(a) Combination 1

(b) Combination 2

(c) Combination 2

Explanation of Solution

a)

Combination 1: $105.5 mL 1.508 M Pb(NO3)2and 250.0 mL 1.2075 M KCl$

Combination 2: $138.5 mL 1.469 M Pb(NO3)2and 100.0 mL 2.115 M KCl$

In Combination 1:

The precipitation reaction between lead nitrate and potassium chloride is as follows:

$2KCl(aq)+Pb(NO3)2(aq)→PbCl2(s)+2KNO3(aq)$

From the above balanced equation, one mole of $Pb(NO3)2$ reacts with two moles of KCl.

First, calculate the number of moles of $Pb(NO3)2$ using the given volume and concentration:

$Moles of Pb(NO3)2= 1.508 M×0.1055 L=0.159094 mol$

Now, calculate the number of moles of KCl.

$Moles of KCl = 1.2075 M×0.250 L=0.301875 mol$

Hence, KCl is a limiting reagent that determines the amount of $PbCl2$ formed.

$Moles of PbCl2 = mol KCl×1 mol PbCl22 mol KCl=0.301875 mol×1 mol PbCl22 mol KCl=0.150938 mol$

Convert moles to grams by multiplying the moles with the molar mass of $PbCl2$

$Mass of PbCl2 = 0.150938 mol×278.1 g/mol PbCl2 = 41.98 g$

Hence, the mass of the $PbCl2$ formed is $41.98 g$ .

In Combination 2,

Calculate the number of moles of $Pb(NO3)2$ using the given volume and concentration

$Moles of Pb(NO3)2 = 1.469 M×0.1385 L=0.203457 mol$

Now, calculate the number of moles of KCl.

$Moles of KCl = 2.115 M×0.100 L=0.2115 mol$

KCl is a limiting reagent that determines the amount of $PbCl2$ formed.

$Moles of PbCl2 = mol KCl×1 mol PbCl22 mol KCl=0.2115 mol×1 mol PbCl22 mol KCl=0.10575 mol$

Now convert the number of moles to grams by multiplying the number of moles with the molar mass of PbCl₂

$g PbCl2= 0.10575 mol×278.1 g/mol PbCl2 = 29.41 g$

Hence, the mass of the PbCl₂ formed is $29.41 g$ .

Therefore, combination 1 will produce a greater mass of the product.

b)

Combination 1: $32.25 mL 0.9475M Na3PO4 and 92.75 mL 0.7750M Ca(NO3)2$

Combination 2: $52.50 mL 0.6810M Na3PO4 and 39.50 mL 1.555M Ca(NO3)2$

Combination 1:

The precipitation reaction between sodium phosphate and calcium nitrate is as follows:

$2Na3PO4(aq)+3Ca(NO3)2(aq)→Ca3(PO4)2(s)+6NaNO3(aq)$

From the above balanced equation, two moles of $Na3PO4$ react with three moles of $Ca(NO3)2$ .

First, calculate the number of moles of Na₃PO₄ using the given volume and concentration:

$Moles of Na3PO4= 0.9475 M×0.03225 L=0.030557 mol$

Now, calculate the number of moles of $Ca(NO3)2$ :

$Moles of Ca(NO3)2= 0.7750 M×0.09275 L=0.071881 mol$

Here, Na₃PO₄ is a limiting reagent that determines the amount of product formed.

$Moles of Ca3(PO4)2 = mol Na3PO4×1 mol Ca3(PO4)22 mol Na3PO4 =0.030557 mol×1 mol Ca3(PO4)22 mol Na3PO4 =0.015279 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of $Ca3(PO4)2$ :

$Mass of Ca3(PO4)2 = 0.015279 mol × 310.18 g/mol Ca3(PO4)2= 4.74 g$

Hence, the mass of the $Ca3(PO4)2$ formed is $4.74 g$ .

Combination 2:

Calculate the number of moles of $Na3PO4$ using the given volume and concentration:

$Moles of Na3PO4 = 0.6810 M × 0.05250 L=0.0357525 mol$

Now, calculate the number of moles of $Ca(NO3)2$ :

$Moles of Ca(NO3)2= 0.7750 M×0.09275 L=0.0614225 mol$

Here, $Na3PO4$ is a limiting reagent that determines the amount of product formed.

$Moles of Ca3(PO4)2 = mol Na3PO4×1 mol Ca3(PO4)22 mol Na3PO4 =0.0357525 mol×1 mol Ca3(PO4)22 mol Na3PO4 =0.01787625 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of $Ca3(PO4)2$ :

$g Ca3(PO4)2= 0.01787625 mol × 310.18 g/mol Ca3(PO4)2= 5.54 g$

Hence, the mass of the $Ca3(PO4)2$ formed is $5.54 g$ .

Therefore, combination 2 produces a greater mass of the product.

c)

Combination 1: $29.75 mL 1.575 M AgNO3 and 25.00 mL 2.010 M BaCl2$

Combination 2: $52.80 mL 2.010 M AgNO3 and 73.50 mL 0.7500 M BaCl2$

Combination 1:

The precipitation reaction between silver nitrate and barium chloride is as follows:

$BaCl2(aq)+2AgNO3(aq)→2AgCl(s)+Ba(NO3)2(aq)$

From the above balanced equation, one mole of BaCl2 reacts with two moles of $AgNO3$ .

First, calculate the number of moles of $BaCl2$ using the given volume and concentration:

$Moles of BaCl2 = 2.010 M×0.0250 L=0.05025 mol$

Now, calculate the number of moles of $AgNO3$ .

$Moles of AgNO3 = 1.575 M×0.02975 L=0.0468563 mol$

Hence, $AgNO3$ is a limiting reagent that determines the amount of AgCl formed.

$Moles of AgCl = mol AgNO3×2 mol AgCl2 mol AgNO3=0.0468563 mol AgNO3×2 mol AgCl2 mol AgNO3=0.0468563 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of AgCl:

$g AgCl = 0.0468563 mol × 143.32 g/mol AgCl = 6.72 g$

Hence, the mass of the AgCl formed is $6.72 g$ .

Combination 2:

Calculate the number of moles of BaCl₂ using the given volume and concentration:

$Moles of BaCl2 = 0.7500 M×0.07350 L=0.055125 mol$

Now, calculate the number of moles of AgNO₃.

$Moles of AgNO3 = 2.010 M×0.05280 L=0.106128 mol$

Hence, $AgNO3$ is a limiting reagent that determines the amount of AgCl formed.

$Moles of AgCl = mol AgNO3×2 mol AgCl2 mol AgNO3=0.106128 mol AgNO3×2 mol AgCl2 mol AgNO3=0.106128 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of AgCl:

$g AgCl = 0.106128 mol×143.32 g/mol AgCl = 15.21 g$

Hence, the mass of the AgCl formed is $15.21 g$ .

Therefore, combination 2 produces a greater mass of the product.

Answer 6

Chapter 4, Problem 91QP

Interpretation Introduction

Interpretation:

The pair of combination that produces a greater mass of the solid product is to be determined.

Concept introduction:

In a precipitation reaction, two ionic compounds in a solution react to form an insoluble compound, which is called the precipitate.

The formation of the precipitate can be predicted by using solubility rules.

The mass of the precipitate can be determined from the stoichiometry of the reaction.

Expert Solution & Answer

Answer to Problem 91QP

Solution:

(a) Combination 1

(b) Combination 2

(c) Combination 2

Explanation of Solution

a)

Combination 1: $105.5 mL 1.508 M Pb(NO3)2and 250.0 mL 1.2075 M KCl$

Combination 2: $138.5 mL 1.469 M Pb(NO3)2and 100.0 mL 2.115 M KCl$

In Combination 1:

The precipitation reaction between lead nitrate and potassium chloride is as follows:

$2KCl(aq)+Pb(NO3)2(aq)→PbCl2(s)+2KNO3(aq)$

From the above balanced equation, one mole of $Pb(NO3)2$ reacts with two moles of KCl.

First, calculate the number of moles of $Pb(NO3)2$ using the given volume and concentration:

$Moles of Pb(NO3)2= 1.508 M×0.1055 L=0.159094 mol$

Now, calculate the number of moles of KCl.

$Moles of KCl = 1.2075 M×0.250 L=0.301875 mol$

Hence, KCl is a limiting reagent that determines the amount of $PbCl2$ formed.

$Moles of PbCl2 = mol KCl×1 mol PbCl22 mol KCl=0.301875 mol×1 mol PbCl22 mol KCl=0.150938 mol$

Convert moles to grams by multiplying the moles with the molar mass of $PbCl2$

$Mass of PbCl2 = 0.150938 mol×278.1 g/mol PbCl2 = 41.98 g$

Hence, the mass of the $PbCl2$ formed is $41.98 g$ .

In Combination 2,

Calculate the number of moles of $Pb(NO3)2$ using the given volume and concentration

$Moles of Pb(NO3)2 = 1.469 M×0.1385 L=0.203457 mol$

Now, calculate the number of moles of KCl.

$Moles of KCl = 2.115 M×0.100 L=0.2115 mol$

KCl is a limiting reagent that determines the amount of $PbCl2$ formed.

$Moles of PbCl2 = mol KCl×1 mol PbCl22 mol KCl=0.2115 mol×1 mol PbCl22 mol KCl=0.10575 mol$

Now convert the number of moles to grams by multiplying the number of moles with the molar mass of PbCl₂

$g PbCl2= 0.10575 mol×278.1 g/mol PbCl2 = 29.41 g$

Hence, the mass of the PbCl₂ formed is $29.41 g$ .

Therefore, combination 1 will produce a greater mass of the product.

b)

Combination 1: $32.25 mL 0.9475M Na3PO4 and 92.75 mL 0.7750M Ca(NO3)2$

Combination 2: $52.50 mL 0.6810M Na3PO4 and 39.50 mL 1.555M Ca(NO3)2$

Combination 1:

The precipitation reaction between sodium phosphate and calcium nitrate is as follows:

$2Na3PO4(aq)+3Ca(NO3)2(aq)→Ca3(PO4)2(s)+6NaNO3(aq)$

From the above balanced equation, two moles of $Na3PO4$ react with three moles of $Ca(NO3)2$ .

First, calculate the number of moles of Na₃PO₄ using the given volume and concentration:

$Moles of Na3PO4= 0.9475 M×0.03225 L=0.030557 mol$

Now, calculate the number of moles of $Ca(NO3)2$ :

$Moles of Ca(NO3)2= 0.7750 M×0.09275 L=0.071881 mol$

Here, Na₃PO₄ is a limiting reagent that determines the amount of product formed.

$Moles of Ca3(PO4)2 = mol Na3PO4×1 mol Ca3(PO4)22 mol Na3PO4 =0.030557 mol×1 mol Ca3(PO4)22 mol Na3PO4 =0.015279 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of $Ca3(PO4)2$ :

$Mass of Ca3(PO4)2 = 0.015279 mol × 310.18 g/mol Ca3(PO4)2= 4.74 g$

Hence, the mass of the $Ca3(PO4)2$ formed is $4.74 g$ .

Combination 2:

Calculate the number of moles of $Na3PO4$ using the given volume and concentration:

$Moles of Na3PO4 = 0.6810 M × 0.05250 L=0.0357525 mol$

Now, calculate the number of moles of $Ca(NO3)2$ :

$Moles of Ca(NO3)2= 0.7750 M×0.09275 L=0.0614225 mol$

Here, $Na3PO4$ is a limiting reagent that determines the amount of product formed.

$Moles of Ca3(PO4)2 = mol Na3PO4×1 mol Ca3(PO4)22 mol Na3PO4 =0.0357525 mol×1 mol Ca3(PO4)22 mol Na3PO4 =0.01787625 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of $Ca3(PO4)2$ :

$g Ca3(PO4)2= 0.01787625 mol × 310.18 g/mol Ca3(PO4)2= 5.54 g$

Hence, the mass of the $Ca3(PO4)2$ formed is $5.54 g$ .

Therefore, combination 2 produces a greater mass of the product.

c)

Combination 1: $29.75 mL 1.575 M AgNO3 and 25.00 mL 2.010 M BaCl2$

Combination 2: $52.80 mL 2.010 M AgNO3 and 73.50 mL 0.7500 M BaCl2$

Combination 1:

The precipitation reaction between silver nitrate and barium chloride is as follows:

$BaCl2(aq)+2AgNO3(aq)→2AgCl(s)+Ba(NO3)2(aq)$

From the above balanced equation, one mole of BaCl2 reacts with two moles of $AgNO3$ .

First, calculate the number of moles of $BaCl2$ using the given volume and concentration:

$Moles of BaCl2 = 2.010 M×0.0250 L=0.05025 mol$

Now, calculate the number of moles of $AgNO3$ .

$Moles of AgNO3 = 1.575 M×0.02975 L=0.0468563 mol$

Hence, $AgNO3$ is a limiting reagent that determines the amount of AgCl formed.

$Moles of AgCl = mol AgNO3×2 mol AgCl2 mol AgNO3=0.0468563 mol AgNO3×2 mol AgCl2 mol AgNO3=0.0468563 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of AgCl:

$g AgCl = 0.0468563 mol × 143.32 g/mol AgCl = 6.72 g$

Hence, the mass of the AgCl formed is $6.72 g$ .

Combination 2:

Calculate the number of moles of BaCl₂ using the given volume and concentration:

$Moles of BaCl2 = 0.7500 M×0.07350 L=0.055125 mol$

Now, calculate the number of moles of AgNO₃.

$Moles of AgNO3 = 2.010 M×0.05280 L=0.106128 mol$

Hence, $AgNO3$ is a limiting reagent that determines the amount of AgCl formed.

$Moles of AgCl = mol AgNO3×2 mol AgCl2 mol AgNO3=0.106128 mol AgNO3×2 mol AgCl2 mol AgNO3=0.106128 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of AgCl:

$g AgCl = 0.106128 mol×143.32 g/mol AgCl = 15.21 g$

Hence, the mass of the AgCl formed is $15.21 g$ .

Therefore, combination 2 produces a greater mass of the product.

Answer 7

Chapter 4, Problem 91QP

Interpretation Introduction

Interpretation:

The pair of combination that produces a greater mass of the solid product is to be determined.

Concept introduction:

In a precipitation reaction, two ionic compounds in a solution react to form an insoluble compound, which is called the precipitate.

The formation of the precipitate can be predicted by using solubility rules.

The mass of the precipitate can be determined from the stoichiometry of the reaction.

Answer 8

Expert Solution & Answer

Answer to Problem 91QP

Solution:

(a) Combination 1

(b) Combination 2

(c) Combination 2

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

a)

Combination 1: $105.5 mL 1.508 M Pb(NO3)2and 250.0 mL 1.2075 M KCl$

Combination 2: $138.5 mL 1.469 M Pb(NO3)2and 100.0 mL 2.115 M KCl$

In Combination 1:

The precipitation reaction between lead nitrate and potassium chloride is as follows:

$2KCl(aq)+Pb(NO3)2(aq)→PbCl2(s)+2KNO3(aq)$

From the above balanced equation, one mole of $Pb(NO3)2$ reacts with two moles of KCl.

First, calculate the number of moles of $Pb(NO3)2$ using the given volume and concentration:

$Moles of Pb(NO3)2= 1.508 M×0.1055 L=0.159094 mol$

Now, calculate the number of moles of KCl.

$Moles of KCl = 1.2075 M×0.250 L=0.301875 mol$

Hence, KCl is a limiting reagent that determines the amount of $PbCl2$ formed.

$Moles of PbCl2 = mol KCl×1 mol PbCl22 mol KCl=0.301875 mol×1 mol PbCl22 mol KCl=0.150938 mol$

Convert moles to grams by multiplying the moles with the molar mass of $PbCl2$

$Mass of PbCl2 = 0.150938 mol×278.1 g/mol PbCl2 = 41.98 g$

Hence, the mass of the $PbCl2$ formed is $41.98 g$ .

In Combination 2,

Calculate the number of moles of $Pb(NO3)2$ using the given volume and concentration

$Moles of Pb(NO3)2 = 1.469 M×0.1385 L=0.203457 mol$

Now, calculate the number of moles of KCl.

$Moles of KCl = 2.115 M×0.100 L=0.2115 mol$

KCl is a limiting reagent that determines the amount of $PbCl2$ formed.

$Moles of PbCl2 = mol KCl×1 mol PbCl22 mol KCl=0.2115 mol×1 mol PbCl22 mol KCl=0.10575 mol$

Now convert the number of moles to grams by multiplying the number of moles with the molar mass of PbCl₂

$g PbCl2= 0.10575 mol×278.1 g/mol PbCl2 = 29.41 g$

Hence, the mass of the PbCl₂ formed is $29.41 g$ .

Therefore, combination 1 will produce a greater mass of the product.

b)

Combination 1: $32.25 mL 0.9475M Na3PO4 and 92.75 mL 0.7750M Ca(NO3)2$

Combination 2: $52.50 mL 0.6810M Na3PO4 and 39.50 mL 1.555M Ca(NO3)2$

Combination 1:

The precipitation reaction between sodium phosphate and calcium nitrate is as follows:

$2Na3PO4(aq)+3Ca(NO3)2(aq)→Ca3(PO4)2(s)+6NaNO3(aq)$

From the above balanced equation, two moles of $Na3PO4$ react with three moles of $Ca(NO3)2$ .

First, calculate the number of moles of Na₃PO₄ using the given volume and concentration:

$Moles of Na3PO4= 0.9475 M×0.03225 L=0.030557 mol$

Now, calculate the number of moles of $Ca(NO3)2$ :

$Moles of Ca(NO3)2= 0.7750 M×0.09275 L=0.071881 mol$

Here, Na₃PO₄ is a limiting reagent that determines the amount of product formed.

$Moles of Ca3(PO4)2 = mol Na3PO4×1 mol Ca3(PO4)22 mol Na3PO4 =0.030557 mol×1 mol Ca3(PO4)22 mol Na3PO4 =0.015279 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of $Ca3(PO4)2$ :

$Mass of Ca3(PO4)2 = 0.015279 mol × 310.18 g/mol Ca3(PO4)2= 4.74 g$

Hence, the mass of the $Ca3(PO4)2$ formed is $4.74 g$ .

Combination 2:

Calculate the number of moles of $Na3PO4$ using the given volume and concentration:

$Moles of Na3PO4 = 0.6810 M × 0.05250 L=0.0357525 mol$

Now, calculate the number of moles of $Ca(NO3)2$ :

$Moles of Ca(NO3)2= 0.7750 M×0.09275 L=0.0614225 mol$

Here, $Na3PO4$ is a limiting reagent that determines the amount of product formed.

$Moles of Ca3(PO4)2 = mol Na3PO4×1 mol Ca3(PO4)22 mol Na3PO4 =0.0357525 mol×1 mol Ca3(PO4)22 mol Na3PO4 =0.01787625 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of $Ca3(PO4)2$ :

$g Ca3(PO4)2= 0.01787625 mol × 310.18 g/mol Ca3(PO4)2= 5.54 g$

Hence, the mass of the $Ca3(PO4)2$ formed is $5.54 g$ .

Therefore, combination 2 produces a greater mass of the product.

c)

Combination 1: $29.75 mL 1.575 M AgNO3 and 25.00 mL 2.010 M BaCl2$

Combination 2: $52.80 mL 2.010 M AgNO3 and 73.50 mL 0.7500 M BaCl2$

Combination 1:

The precipitation reaction between silver nitrate and barium chloride is as follows:

$BaCl2(aq)+2AgNO3(aq)→2AgCl(s)+Ba(NO3)2(aq)$

From the above balanced equation, one mole of BaCl2 reacts with two moles of $AgNO3$ .

First, calculate the number of moles of $BaCl2$ using the given volume and concentration:

$Moles of BaCl2 = 2.010 M×0.0250 L=0.05025 mol$

Now, calculate the number of moles of $AgNO3$ .

$Moles of AgNO3 = 1.575 M×0.02975 L=0.0468563 mol$

Hence, $AgNO3$ is a limiting reagent that determines the amount of AgCl formed.

$Moles of AgCl = mol AgNO3×2 mol AgCl2 mol AgNO3=0.0468563 mol AgNO3×2 mol AgCl2 mol AgNO3=0.0468563 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of AgCl:

$g AgCl = 0.0468563 mol × 143.32 g/mol AgCl = 6.72 g$

Hence, the mass of the AgCl formed is $6.72 g$ .

Combination 2:

Calculate the number of moles of BaCl₂ using the given volume and concentration:

$Moles of BaCl2 = 0.7500 M×0.07350 L=0.055125 mol$

Now, calculate the number of moles of AgNO₃.

$Moles of AgNO3 = 2.010 M×0.05280 L=0.106128 mol$

Hence, $AgNO3$ is a limiting reagent that determines the amount of AgCl formed.

$Moles of AgCl = mol AgNO3×2 mol AgCl2 mol AgNO3=0.106128 mol AgNO3×2 mol AgCl2 mol AgNO3=0.106128 mol$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of AgCl:

$g AgCl = 0.106128 mol×143.32 g/mol AgCl = 15.21 g$

Hence, the mass of the AgCl formed is $15.21 g$ .

Therefore, combination 2 produces a greater mass of the product.

Answer 12

Ch. 4.1 - Prob. 1PPA Ch. 4.1 - Prob. 1PPB Ch. 4.1 - Prob. 1PPC Ch. 4.1 - Prob. 1CP Ch. 4.1 - Soluble molecular compounds are __________. a)...Ch. 4.1 - Which of the following compounds is a weak...Ch. 4.1 - 4.1.4 Which of the following compounds is a strong...Ch. 4.2 - Prob. 1PPA Ch. 4.2 - Prob. 1PPB Ch. 4.2 - Practice Problem CONCEPTUALIZE Using Tables 4.2...

Answer to Problem 91QP

Explanation of Solution

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Chapter 4 Solutions