Chapter 9, Problem 47A

Interpretation Introduction

Interpretation: The expected yield of sodium bicarbonate needs to be calculated under the given reaction conditions.

Concept introduction: In a chemical reaction, the amount of product formed in the relation to the amount of reactant consumed is term as yield.

Answer to Problem 47A

The expected yield of sodium bicarbonate is $28.64\text{ g}$ .

Explanation of Solution

Given:

Chemical reaction for the formation of baking soda, sodium hydrogen carbonate is:

  ${\text{NaCl(aq) + NH}}_{\text{3}}{\text{(aq) + CO}}_{\text{2}}{\text{(s) + H}}_{\text{2}}\text{O(l)}\to {\text{NaHCO}}_{\text{3}}{\text{(s) + NH}}_{\text{4}}\text{Cl(aq)}$

Mass of ${\text{NH}}_{\text{3}}$ = 10 g

Mass of ${\text{CO}}_2$ = 15 g

The number of moles of ${\text{NH}}_{\text{3}}$ and ${\text{CO}}_2$is calculated using formula:

  $\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}$

Since, molar mass of ${\text{NH}}_{\text{3}}$ = 17 g/mol and molar mass of ${\text{CO}}_2$ = 44 g/mol so,

  $\begin{array}{l}{\text{Moles of NH}}_{\text{3}}=\frac{{\text{Mass of NH}}_{\text{3}}}{{\text{Molar mass NH}}_{\text{3}}}\\ {\text{Moles of NH}}_{\text{3}}=\frac{10\text{ g}}{17\text{ g/mol}}\\ {\text{Moles of NH}}_{\text{3}}=0.5882\text{ mol}\end{array}$

  $\begin{array}{l}{\text{Moles of CO}}_{\text{2}}=\frac{{\text{Mass of CO}}_{\text{2}}}{{\text{Molar mass CO}}_{\text{2}}}\\ {\text{Moles of CO}}_{\text{2}}=\frac{15\text{ g}}{44\text{ g/mol}}\\ {\text{Moles of CO}}_{\text{2}}=0.3409\text{ moles}\end{array}$

The limiting reagent is determined as follows:

In the above balanced reaction, 1.0 mol of ${\text{NH}}_{\text{3}}$ reacts with 1.0 mol of ${\text{CO}}_2$ .

There are 0.5582 moles ${\text{NH}}_{\text{3}}$ and 0.3409 moles ${\text{CO}}_2$ . Since the former is present in excess so, ${\text{CO}}_2$ will be the limiting reagent and will determine the amount of ${\text{NaHCO}}_{\text{3}}$ formed.

From the balanced chemical reaction, the mole ratio of ${\text{NaHCO}}_{\text{3}}$ and ${\text{CO}}_2$ is 1:1 so, the number of moles of ${\text{NaHCO}}_{\text{3}}$ formed from 0.3409 moles ${\text{CO}}_2$ is:

  $0.3409{\text{ mol of CO}}_2\text{×}\frac{{\text{1 mol of NaHCO}}_{\text{3}}}{{\text{1 mol of CO}}_2}\text{ = }0.3409{\text{ mol of NaHCO}}_{\text{3}}$

Now, the mass of ${\text{NaHCO}}_{\text{3}}$ produced is calculated as:

The molar mass of ${\text{NaHCO}}_{\text{3}}$ is 84 g/mol. So,

  $\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ 0.3409\text{ mol}=\frac{\text{Mass}}{84\text{ g/mol}}\\ \text{Mass = }0.3409\text{ mol}\times 84\text{ g/mol}\\ \text{Mass = }28.64\text{ g}\end{array}$

Hence, the expected yield of sodium bicarbonate is $28.64\text{ g}$ .

Conclusion

  $28.64\text{ g}$ is the expected yield of sodium bicarbonate.