Chemistry - Stoichiometry Question with Solution | TestHub

${\mathrm{P}}_4{\mathrm{O}}_6+6{\mathrm{H}}_2\mathrm{O}\to 4{\mathrm{H}}_3{\mathrm{PO}}_3..\dots \dots \dots \dots \left(1\right)$

Neutralisation :

$${\mathrm{H}}_3{\mathrm{PO}}_3+2\mathrm{NaOH}.\dots \dots \dots \dots \left(2\right)$$

$\to \left[{\mathrm{Na}}_2{\mathrm{HPO}}_3+2{\mathrm{H}}_2\mathrm{O}\right]\times 4$ Adding eqs. (i) and (ii)

$${\mathrm{P}}_4{\mathrm{O}}_6+8\mathrm{NaOH}\to 4{\mathrm{Na}}_2{\mathrm{HPO}}_3+2{\mathrm{H}}_2\mathrm{O}.\dots \dots \dots \dots \dots \left(3\right)$$

1 mol .8 mol (iii)

Number of moles of ${\mathrm{P}}_4{\mathrm{O}}_6$

$\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}=\frac{1.1}{220}=\frac{1}{200}$ mol.

(molar mass

${\mathrm{P}}_4{\mathrm{O}}_6=(4\times 31)+(6\times 16)=220$

∴ Product formed by 1 mole of ${\mathrm{P}}_4{\mathrm{O}}_6$

is neutralised by 8 moles NaOH

molarity $=\frac{\text{ number of moles }}{\text{ volume in litres }}$

volume $=\frac{\text{ number of moles }}{\text{ molarity }}$

$=\frac{8}{200}\times \frac{1}{0.1}=0.4\mathrm{L}$ or 400 mL

$\therefore 400\mathrm{mL}\mathrm{NaOH}$ is required