Chapter 11, Problem 3STP

Interpretation Introduction

Interpretation:

Using the amount of chemicals available, the maximum amount of moles of NaOH that can be produced needs to be determined.

Concept introduction − Reaction yield depends on the limiting reagent.

Answer to Problem 3STP

(c) Amount of NaOH produces is equal to 8.10 or 8.097 mole

Explanation of Solution

Given:

The LeBlanc process is the traditional method of manufacturing sodium hydroxide. The equation for this process is as follows,

$\underset{}{{\text{Na}}_2{\text{CO}}_3\left(\text{aq}\right)}+\underset{}{{\text{Ca(OH)}}_2\left(\text{aq}\right)}\to 2\text{NaOH}\left(\text{aq}\right){\text{ + CaCO}}_3\left(\text{s}\right)$

$\underset{500\text{g}}{{\text{Na}}_2{\text{CO}}_3}+\underset{300\text{g}}{{\text{Ca(OH)}}_2}\to 2{\text{NaOH + CaCO}}_3$

According to equation one mole of ${\text{Na}}_2{\text{CO}}_3$ and one mole of ${\text{Ca(OH)}}_2$ produces 2 moles of NaOH.

Moles of ${\text{Na}}_2{\text{CO}}_3$ = $\frac{500}{108}=4.629\text{ mole}$

Moles of ${\text{Ca(OH)}}_2$ = $\frac{300}{74}=4.05\text{ mole}$

Since, ${\text{Ca(OH)}}_2$ present in limited amount. So it is limiting reagent.

Thus, NaOH produces is equal to $4.05\times 2=8.10\text{mole}$

Conclusion

Amount of NaOH produces is equal to 8.10 or 8.097 mole.