Question

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

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

If $9.3\mathrm{g}$ of phosphorus was used in the reaction, calculate:

(b) The mass of phosphoric acid formed? $\left[\mathrm{H}=1,\mathrm{N}=14,\mathrm{P}=31,\mathrm{O}=16\right]$

Answer 1

Step 1: Given information

• Phosphorus $\left(\mathrm{P}\right)$ undergoes a chemical reaction with concentrated nitric acid $\left({\mathrm{HNO}}_3\right)$ and results in the formation of phosphoric acid $\left({\mathrm{H}}_3{\mathrm{PO}}_4\right)$, water $\left({\mathrm{H}}_2\mathrm{O}\right)$ and nitrogen dioxide $\left({\mathrm{NO}}_2\right)$.
• The given reaction is as follows:

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

• The given mass of Phosphorus$\left(\mathrm{P}\right)=9.3\mathrm{g}$.
• The atomic mass of Hydrogen$\left(\mathrm{H}\right)=1\mathrm{g}$.
• The atomic mass of Phosphorus$\left(\mathrm{P}\right)=31\mathrm{g}$.
• The atomic mass of Nitrogen$\left(\mathrm{N}\right)=14\mathrm{g}$.
• The atomic mass of Oxygen$\left(\mathrm{O}\right)=16\mathrm{g}$.

Step 2: Calculation of gram molecular mass of Phosphoric acid.

• The gram molecular mass of phosphoric acid is calculated as follows:

$\begin{array}{rcl}\mathrm{Gram}\mathrm{molecular}\mathrm{mass}\mathrm{of}{\mathrm{H}}_3{\mathrm{PO}}_4& =& 1\mathrm{g}\times 3+31\mathrm{g}+16\mathrm{g}\times 4\\ & =& 98\mathrm{g}\end{array}$

Step 3: Calculation of mass of Phosphoric acid produced

Atomic weight of Phosphorus$\left(\mathrm{P}\right)=31\mathrm{g}$.

• Since, $31\mathrm{g}$ of phosphorous produces $98\mathrm{g}$ of phosphoric acid.
• Thus, $9.3\mathrm{g}$ of phosphorous will produce $\frac{98}{31}\times 9.3=29.4\mathrm{g}$ of phosphoric acid.

Therefore, $29.4\mathrm{g}$ of phosphoric acid is formed from $9.3\mathrm{g}$ of phosphorus.