Question

The unbalanced equation for the reaction of $P_4{S}_3$ with nitrate in aqueous acidic medium is given below.
$P_4{S}_3+N{O}_3^{-}\to H_{3}P{O}_4+S{O}_4^{2-}+NO$
The number of moles of water required per mol of $P_4{S}_3$ is $\frac{x}{3}$ , the value of $\text{x}$ is

Answer 1

Answer: 8

$P_4{S}_3+N{O}_3^{-}\to H_{3}P{O}_4+S{O}_4^{2-}+NO$
$\left(P_4{S}_3+28H_{2}O\to 4H_{3}P{O}_4+3S{O}_4^{2-}+44H^{+}+38e^{-}\right)\times 3$
$\left(N{O}_3^{-}+4H^{+}+3e^{-}\to NO+2H_{2}O\right)\times 38$
$\therefore 3P_4{S}_3+38N{O}_3^{-}+8H_{2}O+20H^{+}\to 12H_{3}P{O}_4+9S{O}_4^{2-}+38NO$