Chapter 11, Problem 59A

(a)

Interpretation Introduction

Interpretation:

The balanced chemical equation for the reaction of hydrogen sulfide and oxygen that results the formation of water vapor and sulfur needs to be determined.

Concept introduction:

A balanced chemical equation must represent the molecules of reactant and product in their molecular formulae.

A balanced chemical equation always has same number of all atoms at both sides of the equation and can be used for stoichiometric calculations.

Answer to Problem 59A

${\text{2 H}}_{\text{2}}{\text{S(g) + O}}_{\text{2}}\text{(g) }\stackrel{}{\to }{\text{ 2 H}}_{\text{2}}\text{O(g) + 2 S(s) }$

Explanation of Solution

The reaction of hydrogen sulfide and oxygen forms sulfur and water vapor. The balanced chemical equation can be written as:

${\text{2 H}}_{\text{2}}{\text{S(g) + O}}_{\text{2}}\text{(g) }\stackrel{}{\to }{\text{ 2 H}}_{\text{2}}\text{O(g) + 2 S(s) }$

In the balanced chemical equation, both sides have same number of H, O and S atoms.

(b)

Interpretation Introduction

Interpretation:

The balanced chemical equation for the reaction of hydrogen sulfide and oxygen that results the formation of water vapor and sulfur needs to be determined with the help of given color sketch. (Yellow balls = S, Blue = H, Red = O)

Concept introduction:

A balanced chemical equation must represent the molecules of reactant and product in their molecular formulae.

A balanced chemical equation always has same number of atoms at both sides of the equation and can be use for stoichiometric calculations.

Answer to Problem 59A

Explanation of Solution

The reaction of hydrogen sulfide and oxygen forms sulfur and water vapor. The balanced chemical equation can be written as:

${\text{2 H}}_{\text{2}}{\text{S(g) + O}}_{\text{2}}\text{(g) }\stackrel{}{\to }{\text{ 2 H}}_{\text{2}}\text{O(g) + 2 S(s) }$

In the given figure, 6 moles of H₂S reacts with 3 moles of ${\text{O}}_{\text{2}}$. The mole ratio of reactants can be written as:

2 moles of ${\text{H}}_{\text{2}}\text{S}$ = 1 moles of ${\text{O}}_{\text{2}}$

So 6 moles of ${\text{H}}_{\text{2}}\text{S}$ needs:

${\text{6 moles H}}_{\text{2}}\text{S ×}\frac{{\text{1 mole O}}_{\text{2}}}{{\text{2 moles H}}_{\text{2}}\text{S}}{\text{= 3 moles O}}_{\text{2}}$

$\begin{array}{l}{\text{6 moles H}}_{\text{2}}\text{S ×}\frac{{\text{2 mole H}}_{\text{2}}\text{O}}{{\text{2 moles H}}_{\text{2}}\text{S}}{\text{= 6 moles H}}_{\text{2}}\text{O}\\ {\text{6 moles H}}_{\text{2}}\text{S ×}\frac{\text{2 mole S}}{{\text{2 moles H}}_{\text{2}}\text{S}}\text{= 6 moles S}\end{array}$

Thus, all the 6 moles of ${\text{H}}_{\text{2}}\text{S}$ are consumed to form 3 moles of ${\text{O}}_{\text{2}}$ and also to form 6 moles of each product.