Chapter 9, Problem 9A

(a)

Interpretation Introduction

Interpretation: The mass of the product produced by the complete reaction of 0.125 mol of the first reactant in the given reaction.

  ${\text{AgNO}}_{\text{3}}\text{}\text{(aq)+LiOH(aq)}\to {\text{AgOH(s)+LiNO}}_{\text{3}\text{}}\text{(aq)}$

Concept Introduction:

According to the balanced equation for 1 mol of reaction,

1 mol of AgNO₃ produces 1 mol of AgOH and 1 mol of LiNO₃.

Answer to Problem 9A

15.6 g of AgOH and 8.62 g of LiNO₃ will be produced.

Explanation of Solution

  ${\text{AgNO}}_{\text{3}}\text{}\text{(aq)+LiOH(aq)}\to {\text{AgOH(s)+LiNO}}_{\text{3}\text{}}\text{(aq)}$

According to the balanced equation for 1 mol of reaction,

1 mol of AgNO₃ produces 1 mol of AgOH and 1 mol of LiNO₃

So, 0.125 mol of AgNO₃ produces 0.125 mol of AgOH and 0.125 mol of LiNO₃

Molar mass of AgOH = 124.88 g/mol

Molar mass of LiNO₃ = 68.946 g/mol

  $\text{Number of moles =}\frac{\text{Mass}}{\text{Molar mass}}$

So,

  $\begin{array}{l}massof0.125molofAgOH=\left(0.125mol\times 124.88g/mol\right)=15.6g\\ massof0.125molofLiN{O}_3=\left(0.125mol\times 68.946g/mol\right)=8.62g\end{array}$

(b)

Interpretation Introduction

Interpretation: The mass of the product produced by the complete reaction of 0.125 mol of the first reactant in the given reaction.

  ${\text{Al}}_{\text{2}}\text{}{{\text{(SO}}_{\text{4}}\text{}\text{)}}_{\text{3}}\text{}{\text{(aq)+3CaCl}}_{\text{2}}\text{}\text{(aq)}\to {\text{2AlCl}}_{\text{3}}\text{}{\text{(aq)+3CaSO}}_{\text{4}}\text{}\text{(s)}$

Concept Introduction:

According to the balanced equation for 1 mol of reaction,

1 mol of Al₂(SO₄)₃ produces 2 moles of AlCl₃ and 3 moles of CaSO₄

Answer to Problem 9A

33.3 g of AlCl₃ and 51.1 g of CaSO₄ will be produced.

Explanation of Solution

  ${\text{Al}}_{\text{2}}\text{}{{\text{(SO}}_{\text{4}}\text{}\text{)}}_{\text{3}}\text{}{\text{(aq)+3CaCl}}_{\text{2}}\text{}\text{(aq)}\to {\text{2AlCl}}_{\text{3}}\text{}{\text{(aq)+3CaSO}}_{\text{4}}\text{}\text{(s)}$

According to the balanced equation for 1 mol of reaction,

1 mol of Al₂(SO₄)₃ produces 2 moles of AlCl₃ and 3 moles of CaSO₄

So, 0.125 mol of Al₂(SO₄)₃ produces ?(2×0.125)? mol of AlCl₃ and ?(3×0.125)? mol of CaSO₄

Hence, 0.125 mol of Al₂(SO₄)₃ produces 0.250 mol of AlCl₃ and 0.375 mol of CaSO₄

Molar mass of AlCl₃ = 133.34 g/mol

Molar mass of CaSO₄ = 136.14 g/mol

  $\begin{array}{l}\text{mass of 0}{\text{.250 mol of AlCl}}_{\text{3}}\text{ =}\left(\text{0}\text{.250mol×133}\text{.34g/mol}\right)\text{=33}\text{.3g}\\ \text{mass of 0}{\text{.375 mol of CaSO}}_{\text{4}}\text{ =}\left(\text{0}\text{.375mol×136}\text{.14g/mol}\right)\text{=51}\text{.1g}\end{array}$

(c)

Interpretation Introduction

Interpretation: The mass of the product produced by the complete reaction of 0.125 mol of the first reactant in the given reaction.

  ${\text{CaCO}}_{\text{3}}\text{}\text{(s)+2HCl(aq)}\to {\text{CaCl}}_{\text{2}}\text{}{\text{(aq)+CO}}_{\text{2}}\text{}{\text{(g)+H}}_{\text{2}}\text{}\text{O(l)}$

Concept Introduction:

According to the balanced equation for 1 mol of reaction,

1 mol of CaCO₃ produces 1 mol of CaCl₂, 1 mol of CO₂ and 1 mol of H₂O

Answer to Problem 9A

13.9 g of CaCl₂, 5.50 g of CO₂ and 2.25 g of H₂O will be produced.

Explanation of Solution

? ${\text{CaCO}}_{\text{3}}\text{}\text{(s)+2HCl(aq)}\to {\text{CaCl}}_{\text{2}}\text{}{\text{(aq)+CO}}_{\text{2}}\text{}{\text{(g)+H}}_{\text{2}}\text{}\text{O(l)}$

According to the balanced equation for 1 mol of reaction:

1 mol of CaCO₃ produces 1 mol of CaCl₂, 1 mol of CO₂ and 1 mol of H₂O

So, 0.125 mol of CaCO₃ produces 0.125 mol of CaCl₂, 0.125 mol of CO₂ and 0.125 mol of H₂O

Molar mass of CaCl₂ = 110.98 g/mol

Molar mass of CO₂ = 44.01 g/mol

Molar mass of H₂O = 18.02 g/mol

  $\begin{array}{l}massof0.125molofCaC{l}_2=\left(0.125mol\times 110.98g/mol\right)=13.9g\\ massof0.125molof{H}_{2}O=\left(0.125mol\times 18.02g/mol\right)=2.25g\\ massof0.125molofC{O}_2=\left(0.125mol\times 44.01g/mol\right)=5.50g\\ \end{array}$

(d)

Interpretation Introduction

Interpretation: The mass of the product produced by the complete reaction of 0.125 mol of the first reactant in the given reaction.

  ${\text{2C}}_{\text{4}}\text{}{\text{H}}_{\text{10}\text{}}{\text{(g)+13O}}_{\text{2}}\text{}\text{(g)}\to {\text{8CO}}_{\text{2}}\text{}{\text{(g)+10H}}_{\text{2}}\text{}\text{O(g)}$

Concept Introduction:

According to the balanced equation for 1 mol of reaction:

2 moles of C₄H₁₀ produce 8 moles of CO₂ and 10 moles of H₂O

Answer to Problem 9A

22.0 g of CO₂ and 11.3 g of H₂O will be produced.

Explanation of Solution

  ${\text{2C}}_{\text{4}}\text{}{\text{H}}_{\text{10}\text{}}{\text{(g)+13O}}_{\text{2}}\text{}\text{(g)}\to {\text{8CO}}_{\text{2}}\text{}{\text{(g)+10H}}_{\text{2}}\text{}\text{O(g)}$

According to the balanced equation for 1 mol of reaction:

2 moles of C₄H₁₀ produce 8 moles of CO₂ and 10 moles of H₂O

So,

0.125 mol of C₄H₁₀ produces ?(28?×0.125)? mol of CO₂ and ?(210?×0.125)? mol of H₂O

Hence, 0.125 mol of C₄H₁₀ produces 0.500 mol of CO₂ and 0.625 mol of H₂O

Molar mass of CO₂ = 44.01 g/mol

Molar mass of H₂O = 18.02 g/mol

  $\begin{array}{l}massof0.5molofC{O}_2=\left(0.5mol\times 44.01g/mol\right)=22.0g\\ massof0.625molof{H}_{2}O=\left(0.625mol\times 18.02g/mol\right)=11.3g\end{array}$