Question

Concentrated nitric acid oxidizes phosphorus to phosphoric acid according to the following equation:

$\mathrm{P}+5{\mathrm{HNO}}_3\to {\mathrm{H}}_3{\mathrm{PO}}_4+{\mathrm{H}}_2\mathrm{O}+5{\mathrm{NO}}_2$

(b) What mass of nitric acid will be consumed at the same time?

Answer 1

Given information

• Concentrated nitric acid $\left({\mathrm{HNO}}_3\right)$ oxidizes phosphorus $\left(\mathrm{P}\right)$ to phosphoric acid $\left({\mathrm{H}}_3{\mathrm{PO}}_4\right)$.
• The given reaction is as follows:

$\mathrm{P}+5{\mathrm{HNO}}_3\to {\mathrm{H}}_3{\mathrm{PO}}_4+{\mathrm{H}}_2\mathrm{O}+5{\mathrm{NO}}_2$

• The given mass of Phosphorus$\left(\mathrm{P}\right)=6.2\mathrm{g}$ (From the previous part of the question)
• $19.6\mathrm{g}$ of phosphoric acid can be prepared from $6.2\mathrm{g}$ of phosphorus.

Calculation of molecular weight of Phosphoric acid and nitric acid

$\begin{array}{rcl}\mathrm{Molecular}\mathrm{weight}\mathrm{of}{\mathrm{H}}_3{\mathrm{PO}}_4& =& 3\mathrm{H}+\mathrm{P}+4\mathrm{O}\\ & =& 1\mathrm{g}\times 3+31\mathrm{g}+16\mathrm{g}\times 4\\ & =& 98\mathrm{g}\end{array}$

$\begin{array}{rcl}\mathrm{Molecular}\mathrm{weight}\mathrm{of}5\mathrm{mol}{\mathrm{HNO}}_3& =& 5\times \left(\mathrm{H}+\mathrm{N}+3\mathrm{O}\right)\\ & =& 5\times \left(1\mathrm{g}+14\mathrm{g}+16\mathrm{g}\times 3\right)\\ & =& 315\mathrm{g}\end{array}$

Calculation of mass of nitric acid consumed by stoichiometric calculation

• $31\mathrm{g}$ of phosphorous reacts with $315\mathrm{g}$ of nitric acid.
• $6.2\mathrm{g}$ of phosphorous reacts with $\frac{315}{31}\times 6.2=63\mathrm{g}$ of nitric acid.

Therefore, $63\mathrm{g}$ of nitric acid will be consumed by $6.2\mathrm{g}$ of phosphorus.