Question

When treated with acid, lead (IV) oxide is reduced to a lead (II) salt, with the liberation of oxygen:$\mathbf{2}{\mathbf{PbO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathbf{HNO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathbf{}\mathbf{Pb}{\left({\mathrm{NO}}_3\right)}_{\mathbf{2}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{\ell }\mathbf{\right)}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$.

What volume of a7.91 Msolution of nitric acid is just sufficient to react with15.9gof lead (IV) oxide, according to this equation?

Short answer

The volume of 7.91M solution of nitric acid just sufficient to react with 15.9g of lead (IV) oxide according to the equation is 16.8mL.

Step-by-step solution

The concept of the number of moles

A mole is defined as the mass of the substance which consists of an equal quantity of basic units.One mole of any substance is equal to the value of$\mathbf{6}\mathbf{.}\mathbf{023}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}$(Avogadro number). It can be used to measure the products obtained from the chemical reaction. The unit is denoted by mol.

The formula for the number of moles formula is expressed as:

$\mathbf{n}\mathbf{=}\frac{\mathbf{Mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{substance}}{\mathbf{Mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{one}\mathbf{}\mathbf{mole}}$

Find the number of moles of PbO2

Substitute the values of mass and molar mass of ${\mathrm{PbO}}_2$in the above formula as:

$\begin{array}{l}\mathrm{n}(\mathrm{Pb}{\mathrm{O}}_2)=\frac{15.9\mathrm{g}}{239.2{\mathrm{gmol}}^{-1}}\\ \mathrm{n}(\mathrm{Pb}{\mathrm{O}}_2)=0.0665\mathrm{mol}\end{array}$

Find the number of moles of nitric acid

The number of moles of nitric acid is calculated using stoichiometry as follows:

$\begin{array}{l}\mathrm{n}(\mathrm{HN}{\mathrm{O}}_3)=0.0665\mathrm{mol}{\mathrm{PbO}}_2\times \frac{4{\mathrm{molHNO}}_3}{2{\mathrm{molPbO}}_2}\\ \mathrm{n}(\mathrm{HN}{\mathrm{O}}_3)=0.133\mathrm{mol}\end{array}$

Determine the volume of nitric acid required to react with PbO2

The volume of 7.91M solution of nitric acid just sufficient to react with 15.9g of lead (IV) oxide according to the equation is 16.8mL.