Question

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

$\mathrm{P}+5{\mathrm{HNO}}_3{\left(\mathrm{CONC}.\right)}_{\to }{\mathrm{H}}_3{\mathrm{PO}}_4+{\mathrm{H}}_2\mathrm{O}+5{\mathrm{NO}}_2$

If 9.3g of phosphorus was used in the reactions, calculate: The volume of nitrogen dioxide produced at STP.

Answer 1

Step-1. Given reaction :

$$\begin{gathered} \mathrm{P}\left(\mathrm{s}\right)+5{\mathrm{HNO}}_3(\mathrm{conc}.)\left(\mathrm{L}\right)\to {\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{L}\right)+5{\mathrm{NO}}_2\left(\mathrm{g}\right) \\ \left(\mathrm{Phosphorus}\right)(\mathrm{Nitric}\mathrm{acid})(\mathrm{Phosphoric}(\mathrm{water})(\mathrm{nitrogen} \\ \mathrm{acid})\mathrm{dioxide}) \end{gathered}$$

From the given reaction, it is concluded that 5 moles of ${\mathrm{NO}}_2$ are formed during the reaction.

Step-2. The volume of 1 mole of ${\mathrm{NO}}_2$ at STP $=22.4\times 5$

$=112.0L$

Step-3. If $31\mathrm{g}$ of phosphorus released $112.0\mathrm{L}$ of ${\mathrm{NO}}_2$ at STP

Then $9.3\mathrm{g}$ of Phosphorus releases $=\frac{112}{31}\times 9.3$

$=33.6L$

Hence, $9.3\mathrm{g}$of Phosphorus releases $33.6\mathrm{L}$ of nitrogen dioxide.