Chapter 12.2, Problem 16SP

Interpretation Introduction

Interpretation: The mass of nitrogen dioxide reacting with water to produce $5.00\times {10}^{22}$ molecules of nitrogen monoxide needs to be determined.

Concept Introduction: The relation between the number of moles and the mass of a substance is as follows:

  $n=\frac{m}{M}$

Here, m is mass and M is molar mass.

In 1 mol of a substance, there are $6.022\times {10}^{23}$ units of that substance.

Explanation of Solution

The given reaction is as follows:

  $3{\text{NO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2{\text{HNO}}_{\text{3}}\left(aq\right)+\text{NO}\left(g\right)$

The number of molecules of nitrogen monoxide is $5.00\times {10}^{22}$ molecules.

Since in 1 mol of a substance there are $6.022\times {10}^{23}$ units, thus, the number of moles of nitrogen monoxide corresponding to $5.00\times {10}^{22}$ molecules of nitrogen monoxide will be:

  $n=\frac{1\text{ mol}}{6.022\times {10}^{23}\text{ molecules}}\times 5.00\times {10}^{22}\text{ molecules}$

Or,

  $n=0.083\text{ mol}$

Now, from the given reaction, 3 mol of nitrogen dioxide reacted with water to give 1 mol of nitrogen monoxide. Thus, for 0.083 mol of nitrogen monoxide, the number of moles of nitrogen dioxide required will be:

  $\begin{array}{c}{n}_{{\text{NO}}_{\text{2}}}=\left(\frac{3\text{ mol}}{1\text{ mol}}\right)\left(0.083\text{ mol}\right)\\ =0.25\text{ mol}\end{array}$

Now, the mass of nitrogen dioxide can be calculated from the number of moles and molar mass as follows:

  $m=n\times M$

Put the values,

  $\begin{array}{c}m=\left(0.25\text{ mol}\right)\left(46\text{ g/mol}\right)\\ =11.5\text{ g}\end{array}$

Therefore, the mass of nitrogen dioxide will be 11.5 g.