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 coefficients, $x$ and $y$ are, respectively :

• A. $2$ and $3$
• B. $2$ and $6$
• C. $1$ and $3$
• D. $3$ and $8$

Answer 1

Answer: (B)

$2$ and $6$

The balanced equation is
$Cu⟶{Cu}^{2+}+2e^{-}$ ...(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.