Chapter 9, Problem 50A

Interpretation Introduction

Interpretation:

The theoretical yield of sodium bicarbonate needs to be deduced under the given reaction conditions need to be determined.

Concept Introduction:

• 'Yield refers to the amount of product obtained from a reaction.
• 'Theoretical yield' refers to the amount of product obtained from the molar ratio in the balanced equation whereas the 'actual yield' represents the product recovered from an experimental procedure.
• 'Percent yield' is a ratio of the actual and theoretical yields.

  $\text{\% Yield = }\frac{\text{Actual yield (or experimental yield)}}{\text{Theoretical yield}}\text{×100 ------(1)}$

Answer to Problem 50A

NaHCO₃ = 28.6 g

Explanation of Solution

Given:

Mass of NH₃ = 10 g

Mass of CO₂ = 15 g

Calculations:

The given reaction is:

  ${\text{NaCl + NH}}_{\text{3}}{\text{ + CO}}_{\text{2}}{\text{ + H}}_{\text{2}}\text{O}\to {\text{NaHCO}}_{\text{3}}{\text{ + NH}}_{\text{4}}\text{Cl}$

Step 1: Calculate the moles of NH₃ as follows:

Mass of NH₃ = 10 g

Molar mass of NH₃ = 17 g/mol

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

Step 2: Calculate the moles of CO₂ as follows:

Mass of CO₂ = 15 g

Molar mass of CO₂ = 44 g/mol

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

Step 3: The limiting reagent is determined as follows:

Based on the reaction stoichiometry:

1 mole of NH₃ reacts with 1 mole of CO₂

Here there are 0.5582 moles NH3 and 0.3409 moles CO2. Since the former is present in excess, CO2 will be the limiting reagent and will determine the amount of NaHCO₃ formed.

Step 4: Calculate the moles of NaHCO₃formed as follows:

Based on the reaction stoichiometry:

1 mole of CO₂ forms 1 mole of NaHCO₃

Therefore, 0.3409 moles of CO₂ will produce:

  $\begin{array}{l}\frac{0.3409{\text{ moles CO}}_{\text{2}}\times {\text{ 1 mole NaHCO}}_{\text{3}}}{1{\text{ mole CO}}_{\text{2}}}\\ =0.3409{\text{ moles NaHCO}}_{\text{3}}\end{array}$

Step 5: Calculate the theoretical yield or amount of NaHCO₃

Moles NaHCO₃=0.3409

Molar mass NaHCO₃= 84 g/mol

  $\begin{array}{l}{\text{Mass NaHCO}}_{\text{3}}\text{ = moles }\times \text{ molar mass}\\ \text{=0}\text{.3409}\times \text{84 = 28}\text{.64 g}\end{array}$

Conclusion

Therefore, the yield of sodium bicarbonate is 28.6 g.