Chapter 6, Problem 7PP

Interpretation Introduction

Interpretation:

The limiting reactant and the mass of ${\text{CO}}_2$ formed in the given reaction are to be determined and calculated.

Concept Introduction:

The limiting reactant is the reactant that determines the amount of product that can be formed in a chemical reaction.

The main application of the limiting reactant is that it helps in finding the theoretical yield of a reaction.

Answer to Problem 7PP

Solution:

The limiting reactant is ${\text{Na}}_{\text{2}}{\text{CO}}_3$ and $4.53\text{2 g}$ of ${\text{CO}}_2$ gas is produced when $\text{11 g}$ of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is added to a solution that contains $\text{11 g}$ of $\text{HCl}$ .

Explanation of Solution

Given Information: The given reaction:

${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(s\right)+2\text{HCl}\left(g\right)\to 2\text{NaCl}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)$

Moles of ${\text{Na}}_2{\text{CO}}_3$ is $\text{11 g}$ and moles of $\text{HCl}$ is $\text{11 g}$ .

One mole of ${\text{Na}}_2{\text{CO}}_3$ consists of $106\text{ g}$ of ${\text{Na}}_2{\text{CO}}_3$ . Therefore,

$\begin{array}{c}{\text{106 g Na}}_{\text{2}}{\text{CO}}_3=1{\text{ mole Na}}_{\text{2}}{\text{CO}}_3\\ 1=\frac{1{\text{ mole Na}}_{\text{2}}{\text{CO}}_3}{{\text{106 g Na}}_{\text{2}}{\text{CO}}_3}\end{array}$

For $\text{11 g}$ ,

$\begin{array}{c}{\text{11 g Na}}_{\text{2}}{\text{CO}}_3=\frac{1{\text{ mole Na}}_{\text{2}}{\text{CO}}_3}{{\text{106 g Na}}_{\text{2}}{\text{CO}}_3}\times {\text{11 g Na}}_{\text{2}}{\text{CO}}_3\\ =0.10{\text{3 mole Na}}_{\text{2}}{\text{CO}}_3\end{array}$

One mole of $\text{HCl}$ consists of $36.5\text{ g}$ of $\text{HCl}$ . Therefore,

$\begin{array}{c}\text{36}\text{.5 g HCl}=1\text{ mole HCl}\\ 1=\frac{1\text{ mole HCl}}{\text{36}\text{.5 g HCl}}\end{array}$

For $\text{11 g}$ ,

$\begin{array}{c}\text{11 g HCl}=\frac{1\text{ mole HCl}}{\text{36}\text{.5 g HCl}}\times \text{11 g HCl}\\ =0.301\text{ mole HCl}\end{array}$

From the given reaction, two moles of $\text{HCl}$ react with one mole of ${\text{Na}}_2{\text{CO}}_3$ .

The number of moles of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ required to react with $0.30\text{1 moles}$ of $\text{HCl}$ ,

$\begin{array}{c}\text{0}\text{.301 mol of HCl}=\text{0}\text{.301 mol of HCl}\times \frac{{\text{1 mole of Na}}_{\text{2}}{\text{CO}}_{\text{3}}}{\text{2 mole of HCl}}\\ =\text{0}{\text{.155 mol Na}}_{\text{2}}{\text{CO}}_{\text{3}}\end{array}$

The number of moles required is $0.15\text{5 mol}$ but the actual number of moles that is provided in the given reaction is $0.10\text{3 mol}$ .

Thus, ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is the limiting reactant.

Now, $\text{1 mol}$ of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ gives $\text{1 mol}$ or $\text{44 g}$ of ${\text{CO}}_{\text{2}}$ , then the amount of ${\text{CO}}_{\text{2}}$ produced from $0.10\text{3 mol}$ is :

$\begin{array}{c}1{\text{ mol Na}}_2{\text{CO}}_3=1{\text{ mol CO}}_2\\ 1=\frac{1{\text{ mol CO}}_2}{1{\text{ mol Na}}_2{\text{CO}}_3}\end{array}$

$\begin{array}{c}0.103{\text{ mol Na}}_2{\text{CO}}_3=\frac{1{\text{ mol CO}}_2}{1{\text{ mol Na}}_2{\text{CO}}_3}\times 0.103{\text{ mol Na}}_2{\text{CO}}_3\\ =0.103{\text{ mol CO}}_2\end{array}$

The expression that shows the relation between the moles and the mass of the substance is:

$n=\frac{m}{M}$

Here, $n$ is the number of moles, $M$ is the molar mass, and $m$ is the mass is the substance.

Substitute $0.103\text{ mole}$ for $n$ and $44\text{ g/mol}$ for $M$ in the above expression.

$\begin{array}{c}0.103\text{ mol}=\frac{m}{44\text{ g/mol}}\\ m=4.532\text{ g}\end{array}$

Conclusion

Limiting reactant is ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ and $4.53\text{2 g}$ of ${\text{CO}}_2$ gas is produced.