Chapter 3, Problem 3.93P

Interpretation Introduction

Interpretation:

The percent yield of a reaction in which $45.5\text{ g}$ of ${\text{WO}}_3$ reacts with an excess of ${\text{H}}_2$ gas to produce $\text{W}$ and $9.60\text{ mL}$ of ${\text{H}}_2\text{O}$ is to be determined.

Concept introduction:

Stoichiometry gives the quantitative relationship between the reactants and products of the chemical reaction. It helps in the calculation of the percentage yield of the product of any chemical reaction.

Theoretical yield is defined as the expected amount of product obtained during a reaction. Actual yield is the amount of substance obtained experimentally during the reaction. Percent yield indicates how much of the product is obtained from a reaction. The expression to calculate the percent yield is as follows:

$\text{Percent yield }=\left(\frac{\text{Actual yield}}{\text{Theoretical yield}}\right)\left(\text{100\%}\right)$ (1)

The expression to calculate the amount of a chemical substance in moles is as follows:

$\text{Amount}\left(\text{mol}\right)=\frac{\text{Mass }\left(\text{g}\right)}{\text{Molar mass}\left(\text{g/mol}\right)}$ (2)

The expression to calculate the mass of a chemical substance in grams is as follows:

$\text{Mass}\left(\text{g}\right)=\left(\text{Amount in mol}\right)\left(\text{Molar mass}\left(\text{g/mol}\right)\right)$ (3)

The expression which relates mass, volume and density is as follows:

$\text{Mass}\left(\text{g}\right)=\left(\text{Density}\left(\text{g/mL}\right)\right)\left(\text{Volume}\left(\text{mL}\right)\right)$ (4)