Question

When copper is treated with a certain concentration of nitric acid, nitric oxide and nitrogen dioxide are liberated in equal volumes according to the equation:
$xCu+yHN{O}_3⟶Cu{\left(N{O}_3\right)}_2+NO+N{O}_2+H_{2}O$
The sum of coefficients i.e. $x$ and $y$ is :

Answer 1

The balanced equationa are as follows:
$Cu⟶{Cu}^{2+}+2e^{-}$ ...(i)
$H^{\oplus }+e^{-}+HN{O}_3⟶N{O}_2+H_{2}O$ ...(ii)
$3H^{\oplus }+3e^{-}+HN{O}_3⟶NO+2H_{2}O$ ...(iii)
Multiplying (i) by 2 and adding (i), (ii) and (iii), we get
$2Cu+4H^{\oplus }+2HN{O}_3⟶2{Cu}^{2+}+N{O}_2+NO+3H_{2}O$
or,
$2Cu+6HN{O}_3⟶2Cu{\left(N{O}_3\right)}_2+N{O}_2+NO+3H_{2}O$
$\therefore x$ and $y$ are 2 and 6.
So, $x+y=2+6=8$.