Chapter 9, Problem 7A

Interpretation Introduction

Interpretation:

The mole ratios that would be used to calculate the moles of Ag₂S and H₂ for the given moles of Ag reacting needs to be deduced.

Concept Introduction:

• A chemical reaction is expressed as a chemical equation having reactants and products on left and right side of the reaction arrow respectively.

  $\text{Reactants }\to \text{ Products}$

• The coefficient of a balanced chemical equation i.e. the stoichiometry gives the amount of reactants and products involved in the reaction.
• The stoichiometric ratio between the reactants or between reactants and products is expressed in terms of the mole-mole ratio.

Answer to Problem 7A

  $\begin{array}{l}{\text{Mole ratio for Ag}}_{\text{2}}\text{S }=\frac{1{\text{ mole Ag}}_{\text{2}}\text{S}}{2\text{ moles Ag}}\\ {\text{Mole ratio for H}}_{\text{2}}=\frac{1{\text{ mole H}}_{\text{2}}}{2\text{ moles Ag}}\end{array}$

Explanation of Solution

The given reaction is:

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

The balanced equation is:

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

Based on the reaction stoichiometry:

2 moles of Ag reacts to produce 1 mole of Ag₂S and 1 mole of H₂. Therefore,

Ag:Ag₂S mole-mole ratio = $\frac{1{\text{ mole Ag}}_{\text{2}}\text{S}}{2\text{ moles Ag}}$

Ag:H₂mole-mole ratio = $\frac{1{\text{ mole H}}_{\text{2}}}{2\text{ moles Ag}}$

Conclusion

Therefore, the mole ratio factor is ½ or 0.5 for both Ag₂S and H₂.