Question

Sodium nitrate on reduction with $Zn$ in presence of $NaOH$ solution produces $N{H}_3$ Mass of sodium nitrate absorbing $1$ mole of electron will be

• A. 7.750
• B. 10.625
• C. 8.000
• D. 9.875

Answer 1

Answer: (B)

Required equation is given below,
$Zn+2O{H}^{-}⟶Zn{O}_2^{2-}+2H^{+}+2e^{-}$
$N{O}_3^{-}+8H^{+}+8e^{-}⟶O{H}^{-}+2H_{2}O+N{H}_3$
From the above equation
$\because 8$ moles of electron absorbed by $85g$ of $NaN{O}_3$
$\therefore 1$ mole of electron absorbed by $\frac{85}{8}g$ of $NaN{O}_3$
$=10.625g$