Chapter 3, Problem 94QAP

A 100.0-g mixture made up of NaCl0₃, Na₂CO₃, NaCl, and NaHCO₃ is heated, producing 5.95 g of oxygen, 1.67 g of water, and 14.5 g of carbon dioxide. NaCl does not react under the conditions of the experiment. The equations for the reactions that take place are:

${\text{2NaClO}}_{\text{3}}\left(s\right)\to \text{2NaCl}\left(s\right)+{\text{3O}}_{\text{2}}\left(g\right)$ ${\text{Na}}_2{\text{CO}}_{\text{3}}\left(s\right)\to {\text{2Na}}_2\text{O}\left(s\right)+{\text{CO}}_{\text{2}}\left(g\right)$ ${\text{2NaHCO}}_{\text{3}}\left(s\right)\to {\text{Na}}_2\text{O}\left(s\right)+{\text{2CO}}_{\text{2}}\left(g\right)+{\text{H}}_2\text{O}$

Assuming 100% decomposition of NaClO₃, Na₂CO₃, and NaHCO₃, what is the composition of the mixture in grams?

Interpretation Introduction

Interpretation:

The composition of the mixture should be determined in grams..

Concept introduction:.

The number of moles of a substance is related to mass and molar mass as follows:.

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

Here,mis mass andMis molar mass of the substance..

From a balanced chemical reaction, the number of moles of 1 reactant reacting with number of moles of 2^nd reactant can be calculating using the unitary method.

Answer to Problem 94QAP

The composition of mixture is 13.2 g of ${\text{NaClO}}_{\text{3}}$, 15.6 g of ${\text{NaHCO}}_{\text{3}}$, 15.3 g of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ and 55.9 g of NaCl.

Explanation of Solution

The mass of the mixture is 100.0 g.

The composition of the mixture is ${\mathrm{NaClO}}_3$, ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$, NaCl and ${\text{NaHCO}}_{\text{3}}$..

The mixture produces 5.95 g of oxygen, 1.67 g of water and 14.5 g of carbon dioxide on heating.

The equations for the reactions are as follows:.

$\begin{array}{l}2{\text{NaClO}}_{\text{3}}\left(s\right)\to 2\text{NaCl}\left(s\right)+3{\text{O}}_{\text{2}}\left(g\right)\\ {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(s\right)\to {\text{Na}}_{\text{2}}\text{O}\left(s\right)+{\text{CO}}_{\text{2}}\left(g\right)\\ 2{\text{NaHCO}}_{\text{3}}\left(s\right)\to {\text{Na}}_{\text{2}}\text{O}\left(s\right)+2{\text{CO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\end{array}$.

From the given mass of oxygen, carbon dioxide and water, number of moles can be calculated as follows:.

$n_{O_2}=\frac{m_{O_2}}{M_{O_2}}$.

Molar mass of oxygen gas is 32 g/mol, putting the values,

$n_{O_2}=\frac{5.95\text{ g}}{32\text{ g/mol}}=0.186\text{ mol}$.

Similarly, number of moles of water will be:.

$n_{H_{2}O}=\frac{m_{H_{2}O}}{M_{H_{2}O}}$.

Molar mass of water is 18 g/mol, putting the values,

$n_{H_{2}O}=\frac{1.67\text{ g}}{18\text{ g/mol}}=0.0928\text{ mol}$.

Now, number of moles of carbon dioxide can be calculated as follows:.

$n_{C{O}_2}=\frac{m_{C{O}_2}}{M_{C{O}_2}}$.

Molar mass of carbon dioxide is 44.01 g/mol, putting the values,

$n_{C{O}_2}=\frac{14.5\text{ g}}{44.01\text{ g/mol}}=0.33\text{ mol}$.

From the reaction as follows:.

$2{\text{NaClO}}_{\text{3}}\left(s\right)\to 2\text{NaCl}\left(s\right)+3{\text{O}}_{\text{2}}\left(g\right)$.

3 mol of oxygen gas are produced from 2 mol of ${\text{NaClO}}_{\text{3}}$ or 1 mol of oxygen gas produced from 2/3 mol of ${\text{NaClO}}_{\text{3}}$ thus, number of moles of ${\text{NaClO}}_{\text{3}}$ required for 0.186 mol of oxygen gas will be:.

$n_{{\text{NaClO}}_{\text{3}}}=0.186\times \frac{2}{3}\text{ mol}=0.124\text{ mol}$.

Now, molar mass of ${\text{NaClO}}_3$ is 106.44 g/mol thus, mass of ${\text{NaClO}}_3$ can be calculated as follows:.

$m=n\times M$

Putting the values,

$m=0.124\text{ mol}\times 106.44\text{ g/mol=13}\text{.2 g}$.

Now, from the reaction given below:.

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

1 mol of water is produced from 2 mol of ${\text{NaHCO}}_{\text{3}}$, the number of moles of ${\text{NaHCO}}_{\text{3}}$ required to produce 0.0928 mol of water will be:.

$n_{{\text{NaHCO}}_{\text{3}}}=2\times 0.0928\text{ mol}=0.1856\text{ mol}$.

The molar mass of ${\text{NaHCO}}_{\text{3}}$ is 84.007 g/mol thus, mass of ${\text{NaHCO}}_{\text{3}}$ will be:.

$m=0.1856\text{ mol}\times 84.007\text{ g/mol=15}\text{.6 g}$.

From the calculated number of moles of ${\text{NaHCO}}_{\text{3}}$, the number of moles of carbon dioxide produced can be calculated as follows:.

2 mol of ${\text{NaHCO}}_{\text{3}}$ produces 2 mol of carbon dioxide, thus, 0.1856 mol of ${\text{NaHCO}}_{\text{3}}$ will produce 0.1856 mol of carbon dioxide.

The total number of moles of carbon dioxide is 0.33 mol thus, remaining number of moles of carbon dioxide will be:.

$n_{{\text{CO}}_{\text{2}}\text{,remaining}}=\left(0.33-0.1856\right)\text{ mol}=0.144\text{ mol}$.

The decomposition of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is as follows:.

${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(s\right)\to {\text{Na}}_{\text{2}}\text{O}\left(s\right)+{\text{CO}}_{\text{2}}\left(g\right)$.

From the above reaction, 1 mol of carbon dioxide is produced from 1 mol of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ thus, 0.144 mol of carbon dioxide will be produced from 0.144 mol of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$..

The molar mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is 105.98 g/mol thus, mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ can be calculated as follows:.

$m=0.144\text{ mol}\times 105.98\text{ g/mol=15}\text{.3 g}$.

Now, total mass is 100 g thus, mass of NaCl can be calculated as follows:.

$m_{\text{NaCl}}=100\text{ g}-m_{{\text{NaClO}}_3}-m_{{\text{NaHCO}}_{\text{3}}}-m_{{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}}$.

Putting the values,

$m_{\text{NaCl}}=\left(100-13.2-15.6-15.3\right)\text{g}=\text{55}\text{.9 g}$

Conclusion

Therefore, the composition of mixture is 13.2 g of ${\text{NaClO}}_{\text{3}}$, 15.6 g of ${\text{NaHCO}}_{\text{3}}$, 15.3 g of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ and 55.9 g of NaCl.