Chapter 3, Problem 3.49P

(a)

Interpretation Introduction

Interpretation: Moles of ${\text{H}}_2$ formed form $\text{5}{\text{.70 g SrH}}_2$ are to be calculated.

Concept introduction: The formula to convert mass in grams to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$

For some stoichiometric balanced equation; reactant that is present in the smallest amount is consumed completely to form the product and thus governs the product formation.This reactant is known as limiting reactant.On the basis of the stoichiometric ratio by which substrates combine, the moles of product formed from each reactant can be calculated.

(b)

Interpretation Introduction

Interpretation:Moles of ${\text{H}}_2$ formed form $\text{4}{\text{.75 g H}}_2\text{O}$ are to be calculated.

Concept introduction: The formula to convert mass in grams to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$

The formula to convert moles to mass in grams is as follows:

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{molar mass}\right)$

(c)

Interpretation Introduction

Interpretation: Limiting reactant should be identified.

Concept introduction: The formula to convert mass in grams to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$

The formula to convert moles to mass in grams is as follows:

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{molar mass}\right)$

(d)

Interpretation Introduction

Interpretation: Mass of ${\text{H}}_2$ produced is to be calculated.

Concept introduction: The formula to convert mass in grams to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$

The formula to convert moles to mass in grams is as follows:

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{molar mass}\right)$