Chapter 9, Problem 8A

(a)

Interpretation Introduction

Interpretation:

For the balanced chemical equations, the number of moles of product produced if 0.500 mole of first reactant were to react completely needs to be determined.

  $C{O}_{2(g)}+4H_{2(g)}\to C{H}_{4(g)}+2H_{2}O(l)$

Concept Introduction:

The balanced chemical equation means , 1 mole of CO₂requires 4 moles of H₂ to produce 1 mole of methane and 2 moles of water.

Answer to Problem 8A

0.500 mole or half mole of CO₂ will give half mole i.e. 0.500 mole of CH₄ and 1 mole of H₂O.

Explanation of Solution

  $C{O}_{2(g)}+4H_{2(g)}\to C{H}_{4(g)}+2H_{2}O(l)$

  $\begin{array}{l}0.500{\text{ moles of CO}}_2\times \frac{1{\text{ mole of CH}}_4}{1{\text{ mol of CO}}_2}=0.500{\text{ mol of CH}}_4\\ 0.500{\text{ moles of CO}}_2\times \frac{{\text{2mole of H}}_{2}O}{1{\text{ mol of CO}}_2}=1.00{\text{ mole of H}}_{2}O\end{array}$

(b)

Interpretation Introduction

Interpretation:

For the balanced chemical equations, the number of moles of product produced if 0.500 mole of first reactant were to react completely needs to be determined.

  $BaC{l}_2(aq)+2AgN{O}_3(aq)\to 2AgCl(s)+Ba{(N{O}_3)}_2(aq)$

Concept Introduction:

In the given balanced chemical equation, 1mole of barium chloride require 2 moles of AgNO₃ to produce 2 mole of AgCl and 1 mole of barium nitrate.

Answer to Problem 8A

0.500 mole or half mole of BaCl₂will give half mole i.e. 0.500 mole of Ba(NO₃)₂ and 1 mole of AgCl.

Explanation of Solution

  $BaC{l}_2(aq)+2AgN{O}_3(aq)\to 2AgCl(s)+Ba{(N{O}_3)}_2(aq)$

  $\begin{array}{l}0.500{\text{ moles of BaCl}}_2\times \frac{\text{2mol of }AgCl}{1mole{\text{ of BaCl}}_2}=1.00\text{ mole of AgCl}\\ \frac{1{\text{ mol of Ba(NO}}_3{)}_2}{1mole{\text{ of BaCl}}_2}\times 0.500{\text{ mole of BaCl}}_2=0.500{\text{mol of Ba(NO}}_3{)}_2\end{array}$

(c)

Interpretation Introduction

Interpretation:

For the balanced chemical equations, the number of moles of product produced if 0.500 mole of first reactant were to react completely needs to be determined.

  $C_3{H}_8(g)+5O_2(g)\to 4H_{2}O(l)+3C{O}_2(g)$

Concept Introduction:

In the given balanced chemical equation 1 mol of C₃H₈ require 5 mole of oxygen molecule to produce 4 mole of water and 3 mole of carbon dioxide.

Answer to Problem 8A

0.500 mole or half mole of C₃H₈ will give 2 mole of H₂O and 1.5 mole of CO₂.

Explanation of Solution

  $C_3{H}_8(g)+5O_2(g)\to 4H_{2}O(l)+3C{O}_2(g)$

  $\begin{array}{l}0.500{\text{ moles of C}}_3{H}_8\times \frac{{\text{4mol of H}}_{2}O}{1mole{\text{ of C}}_2{H}_8}=2.00{\text{ mole of H}}_{2}O\\ \frac{{\text{3 mol of CO}}_2}{1mole{\text{ of C}}_3{H}_8}\times 0.500{\text{ mole of C}}_3{H}_8=1.50{\text{mol of CO}}_2\end{array}$

(d)

Interpretation Introduction

Interpretation:

For the balanced chemical equations, the number of moles of product produced if 0.500 mole of first reactant were to react completely needs to be determined.

  $3H_{2}S{O}_4(aq)+2Fe(s)\to F{e}_2{(S{O}_4)}_3(aq)+3H_2(g)$

Concept Introduction:

in the given balanced chemical equation 3 moles of sulphuric acid requires 2 moles of iron to produce 1 mol of ferric sulfate and 3 mole of Hydrogen molecule.

Answer to Problem 8A

0.500 mole or half mole of H₂SO₄ will give 0.167 mole of Fe₂(SO₄)₃and 0.500 mole of H₂

Explanation of Solution

  $3H_{2}S{O}_4(aq)+2Fe(s)\to F{e}_2{(S{O}_4)}_3(aq)+3H_2(g)$

  $\begin{array}{l}0.500{\text{ moles of H}}_{2}S{O}_4\times \frac{{\text{1mol of Fe}}_2{(S{O}_4)}_3}{3mole{\text{ of H}}_{2}S{O}_4}=0.167{\text{ mole of Fe}}_2{(}_S\\ \frac{{\text{3 mol of H}}_2}{3mole{\text{ of H}}_{2}S{O}_4}\times 0.500{\text{ mole of H}}_{2}S{O}_4=0.500{\text{mol of H}}_2\end{array}$