Chapter 4, Problem 4.15PAE

4.15 What mass of the unknown compound is formed in the following reactions? (Assume that reactants for which amounts are not given are present in excess.)

   ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}+{\text{ H}}_{\text{2}} \to {\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$

g C₂H₄ reacts to produce x g C₂H₆

   ${\text{CS}}_{\text{2}}+{\text{ 3Cl}}_{\text{2}} \to {\text{CCl}}_{\text{4}}+{\text{ S}}_{\text{2}}{\text{Cl}}_{\text{2}}$

5.78 g Cl₂ reacts to produce x g S₂CL₂

   ${\text{PCl}}_{\text{3}}+{\text{ 3H}}_{\text{2}}\text{O} \to {\text{H}}_{\text{3}}{\text{PO}}_{\text{3}}+\text{ 3HCl}$

mg PCl₃ reacts to produce x mg HCl

   ${\text{B}}_{\text{2}}{\text{H}}_{\text{6}}+{\text{ O}}_{\text{2}} \to {\text{B}}_{\text{2}}{\text{O}}_{\text{3}}+{\text{ 3H}}_{\text{2}}\text{O}$

kg B₂H₆ reacts to produce x kg B₂O₃

Interpretation Introduction

To determine: The mass of unknown compound that is formed in: ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{+ H}}_{\text{2}}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{6}},\text{ 3}{\text{.4 g C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{ reacts to produce x g C}}_{\text{2}}{\text{H}}_{\text{6}}$.

Explanation of Solution

Calculate the molecular weight of C₂H₄ and C₂H₆

$\begin{array}{l}{\text{MW(C}}_{\text{2}}{\text{H}}_{\text{4}}\text{)=2 }\times \text{ 12 }\frac{\text{g}}{\text{mol}}\text{ + 4 }\times \text{ 1 }\frac{\text{g}}{\text{mol}}=28\frac{\text{g}}{\text{mol}}\\ {\text{MW(C}}_{\text{2}}{\text{H}}_6\text{)=2 }\times \text{ 12 }\frac{\text{g}}{\text{mol}}\text{ + 6 }\times \text{ 1 }\frac{\text{g}}{\text{mol}}=30\frac{\text{g}}{\text{mol}}\end{array}$

1 mol of C₂H₄ produces 1 mol of C₂H₆. Convert these moles in grams

$\begin{array}{l}{\text{1 mol C}}_{\text{2}}{\text{H}}_4\times \frac{{\text{28 g C}}_{\text{2}}{\text{H}}_4}{{\text{1 mol C}}_{\text{2}}{\text{H}}_4}={\text{28 g C}}_{\text{2}}{\text{H}}_4\\ 1{\text{ mol C}}_{\text{2}}{\text{H}}_6\times \frac{{\text{30 g C}}_{\text{2}}{\text{H}}_6}{{\text{1 mol C}}_{\text{2}}{\text{H}}_6}={\text{30 g C}}_{\text{2}}{\text{H}}_6\end{array}$

Calculate the mass of C₂H₆produce with 3.4 g of C₂H₄

$\begin{array}{l}{\text{28 g C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{-----30 g C}}_{\text{2}}{\text{H}}_{\text{6}}\\ \text{3}{\text{.4 g C}}_{\text{2}}{\text{H}}_{\text{4}}\text{----}\frac{\text{3}{\text{.4 g C}}_{\text{2}}{\text{H}}_{\text{4}}\times {\text{30 g C}}_{\text{2}}{\text{H}}_{\text{6}}}{{\text{28 g C}}_{\text{2}}{\text{H}}_{\text{4}}}=3.{\text{6 g C}}_{\text{2}}{\text{H}}_{\text{6}}\end{array}$

Interpretation Introduction

To determine: The mass of unknown compound that is formed in: ${\text{CS}}_{\text{2}}{\text{+ 3Cl}}_{\text{2}}\to {\text{CCl}}_{\text{4}}{\text{ + S}}_{\text{2}}{\text{Cl}}_{\text{2}},\text{ 5}{\text{.78 g Cl}}_{\text{2}}{\text{ reacts to produce x g S}}_{\text{2}}{\text{Cl}}_{\text{2}}$.

Explanation of Solution

Calculate the molecular weight of Cl₂ and S₂Cl₂

$\begin{array}{l}{\text{MW(Cl}}_{\text{2}}\text{)=2 }\times \text{ 35}\text{.5 }\frac{\text{g}}{\text{mol}}=71\frac{\text{g}}{\text{mol}}\\ {\text{MW(S}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{)=2 }\times \text{32 }\frac{\text{g}}{\text{mol}}\text{ + 2 }\times \text{ 35}\text{.5 }\frac{\text{g}}{\text{mol}}=135\frac{\text{g}}{\text{mol}}\end{array}$

3 moles of Cl₂ produces 1 mol of S₂Cl₂. Convert these moles in grams

$\begin{array}{l}{\text{3 moles Cl}}_{\text{2}}\times \frac{{\text{71 g Cl}}_{\text{2}}}{{\text{1 mol Cl}}_{\text{2}}}=213{\text{ g Cl}}_{\text{2}}\\ 1{\text{ mol S}}_{\text{2}}{\text{Cl}}_{\text{2}}\times \frac{{\text{135 g S}}_{\text{2}}{\text{Cl}}_{\text{2}}}{{\text{1 mol S}}_{\text{2}}{\text{Cl}}_{\text{2}}}=135{\text{ g S}}_{\text{2}}{\text{Cl}}_{\text{2}}\end{array}$

Calculate the mass of S₂Cl₂produce with 5.78 g of Cl₂

$\begin{array}{l}{\text{213 g Cl}}_{\text{2}}{\text{-----135 g S}}_{\text{2}}{\text{Cl}}_{\text{2}}\\ \text{5}{\text{.78 g Cl}}_{\text{2}}\text{----}\frac{\text{5}{\text{.78 g Cl}}_{\text{2}}\times {\text{135 g S}}_{\text{2}}{\text{Cl}}_{\text{2}}}{{\text{213 g Cl}}_{\text{2}}}=3.{\text{66 g S}}_{\text{2}}{\text{Cl}}_{\text{2}}\end{array}$

Interpretation Introduction

To determine: The mass of unknown compound that is formed in: ${\text{PCl}}_3{\text{+ 3 H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{3}}{\text{PO}}_{\text{3}}\text{ + 3 HCl},\text{ 3}{\text{.1 mg PCl}}_3\text{ reacts to produce x mg HCl}$.

Explanation of Solution

Given the following reaction,

${\text{ PCl}}_3{\text{+ 3 H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{3}}{\text{PO}}_{\text{3}}\text{ + 3 HCl}$

Calculate the molecular weight of PCl₃ and HCl

$\begin{array}{l}{\text{MW(PCl}}_3\text{)=1 }\times \text{ 31 }\frac{\text{g}}{\text{mol}}\text{+3 }\times \text{ 35}\text{.5 }\frac{\text{g}}{\text{mol}}=137.5\frac{\text{g}}{\text{mol}}\\ \text{MW(HCl)=1 }\times \text{1 }\frac{\text{g}}{\text{mol}}\text{ + 1 }\times \text{35}\text{.5 }\frac{\text{g}}{\text{mol}}=36.5\frac{\text{g}}{\text{mol}}\end{array}$

1 mol of PCl₃ produces 3 moles of HCl. Convert these moles in mg

$\begin{array}{l}{\text{1 moles PCl}}_3\times \frac{\text{137}{\text{.5 g PCl}}_3}{{\text{1 mol PCl}}_3}\times \frac{\text{1000 mg}}{\text{1 g}}=137500{\text{ mg PCl}}_3\\ \text{3 moles HCl }\times \frac{\text{36}\text{.5 g HCl}}{\text{1 mol HCl }}\times \frac{\text{1000 mg}}{\text{1 g}}=109500\text{ mg HCl}\end{array}$

Calculate the mass of HCl produce with 3.1 mg of HCl

$\begin{array}{l}{\text{137500 mg PCl}}_3\text{-----109500 mg HCl}\\ \text{3}{\text{.1 mg PCl}}_3\text{----}\frac{\text{3}{\text{.1 mg PCl}}_3\times \text{109500 mg HCl}}{{\text{137500 mg PCl}}_3}=2.5\text{ mg HCl}\end{array}$

Interpretation Introduction

To determine: The mass of unknown compound that is formed in: ${\text{B}}_{\text{2}}{\text{H}}_{\text{6}}{\text{+ 3 O}}_{\text{2}}\to {\text{B}}_{\text{2}}{\text{O}}_{\text{3}}{\text{ + 3 H}}_{\text{2}}\text{O},\text{ 4}{\text{.6 kg B}}_{\text{2}}{\text{H}}_{\text{6}}{\text{ reacts to produce x kg B}}_{\text{2}}{\text{O}}_{\text{3}}$.

Explanation of Solution

Calculate the molecular weight of B₂H₆ and B₂O₃

$\begin{array}{l}{\text{MW(B}}_{\text{2}}{\text{H}}_{\text{6}}\text{)=2 }\times \text{ 11 }\frac{\text{g}}{\text{mol}}\text{+6}\times \text{ 1 }\frac{\text{g}}{\text{mol}}=28\frac{\text{g}}{\text{mol}}\\ {\text{MW(B}}_{\text{2}}{\text{O}}_{\text{3}}\text{)=2 }\times \text{ 11 }\frac{\text{g}}{\text{mol}}\text{ + 3 }\times \text{16 }\frac{\text{g}}{\text{mol}}=70\frac{\text{g}}{\text{mol}}\end{array}$

1 mol of B₂H₆ produces 1 mol of B₂O₃. Convert these moles in kg

$\begin{array}{l}{\text{1 moles B}}_2{\text{H}}_6\times \frac{{\text{28 g B}}_2{\text{H}}_6}{{\text{1 mol B}}_2{\text{H}}_6}\times \frac{\text{1 kg}}{\text{1000 g}}=0.028{\text{ kg B}}_2{\text{H}}_6\\ {\text{1 moles B}}_2{\text{O}}_{\text{3}}\times \frac{{\text{70 g B}}_2{\text{O}}_{\text{3}}}{{\text{1 mol B}}_2{\text{O}}_{\text{3}}}\times \frac{\text{1 kg}}{\text{1000 g}}=0.070{\text{ kg B}}_2{\text{O}}_{\text{3}}\end{array}$

Calculate the mass of B₂O₃produce with 4.6 kg of B₂H₆

$\begin{array}{l}\text{0}{\text{.028 kg B}}_{\text{2}}{\text{H}}_{\text{6}}\text{-----0}{\text{.070 kg B}}_{\text{2}}{\text{O}}_{\text{3}}\\ \text{4}{\text{.6 kg B}}_{\text{2}}{\text{H}}_{\text{6}}\text{--------}\frac{\text{4}{\text{.6 kg B}}_{\text{2}}{\text{H}}_{\text{6}}\times \text{0}{\text{.070 kg B}}_{\text{2}}{\text{O}}_{\text{3}}}{\text{0}{\text{.028 kg B}}_{\text{2}}{\text{H}}_{\text{6}}}=11.5{\text{ kg B}}_{\text{2}}{\text{O}}_{\text{3}}\end{array}$

Conclusion

a) 3.4 g of C₂H₄ produces $3.{\text{6 g of C}}_{\text{2}}{\text{H}}_{\text{6}}$.

b) 5.78 g of Cl₂ produces $\text{3}{\text{.66 g of S}}_{\text{2}}{\text{Cl}}_{\text{2}}$.

c) 3.1 mg of PCl₃ produces $\text{2}\text{.5 mg of HCl}$.

d) 4.6 kg of B₂H₆ produces $\text{11}{\text{.5 kg of B}}_{\text{2}}{\text{O}}_{\text{3}}$.