Question

Nitrogen oxide can be generated on a laboratory scale by the reaction of dilute sulphuric acid with aqueous sodium nitrite:$\mathbf{6}{\mathbf{NaNO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\to }\mathbf{4}\mathbf{NO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{HNO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{\ell }\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$.

What volume of 0.646Maqueous${\mathbf{NaNO}}_{\mathbf{2}}$should be used in this reaction to generate 5.00Lof nitrogen oxide at a temperature ofand a pressure of0.970atm?

Short answer

The volume of 0.646Maqueous${\mathrm{NaNO}}_2$that should be used in the reaction to generate 5.00L of nitrogen oxide at a temperature of${20}^{0}C$ and a pressure of 0.970atmis 469mL.

Step-by-step solution

The concept of the ideal gas equation

The ideal gas equation is given as: PV=nRT .

Where p is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

Calculate the number of moles of nitrogen oxide

The number of moles is calculated with the use of the ideal equation as: $n=\frac{PV}{RT}$

Substitute the values in this equation to find the number of moles of nitrogen oxide as:

$\begin{array}{l}\mathrm{n}=\frac{0.970\mathrm{atm}\times 5.00\mathrm{L}}{0.08206\mathrm{L}\mathrm{atm}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\times 293\mathrm{K}}\\ \mathrm{n}=0.202\mathrm{mol}\end{array}$

Calculate the number of moles of sodium nitrite

The number of moles ${\mathrm{NaNO}}_2$is calculated by stoichiometry as follows

$\begin{array}{l}\mathrm{n}(\mathrm{NaN}{\mathrm{O}}_2)=0.202\mathrm{mol}\mathrm{NO}\times \frac{6\mathrm{mol}{\mathrm{NaNO}}_2}{4\mathrm{mol}\mathrm{NO}}\\ \mathrm{n}(\mathrm{NaN}{\mathrm{O}}_2)=0.303\mathrm{mol}\end{array}$

Calculate the volume of sodium nitrite required 

The volume of ${\mathrm{NaNO}}_2$required to generate nitrogen oxide is calculated as

$\begin{array}{l}\mathrm{V}(\mathrm{NaN}{\mathrm{O}}_2)=0.303\mathrm{mol}{\mathrm{NaNO}}_2\times \frac{\mathrm{L}}{0.646\mathrm{mol}{\mathrm{NaNO}}_2}\\ \mathrm{V}(\mathrm{NaN}{\mathrm{O}}_2)=0.469\mathrm{L}\\ \mathrm{V}(\mathrm{NaN}{\mathrm{O}}_2)=469\mathrm{mL}\end{array}$

Thus, the volume of 0.646M aqueous${\mathrm{NaNO}}_2$ that should be used in the reaction to generate 5.00L of nitrogen oxide at a temperature${20}^{0}C$ of and a pressure of 0.970atm is 469mL.