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 of 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)

Balanced equation for producing $NO$ and $N{O}_2$ respectively are
$3Cu+8HN{O}_3\to 3Cu{\left(N{O}_3\right)}_2+2NO+4H_{2}O\dots$ (i)
$Cu+4HN{O}_3\to Cu{\left(N{O}_3\right)}_2+2N{O}_2+2H_{2}O\dots$ (ii)
Here $NO$ and $N{O}_2$ are evolved in equal volumes, hence, on adding Eqs. (i) and (ii)
$4Cu+12HN{O}_3\to 4Cu{\left(N{O}_3\right)}_2+2NO+2N{O}_2+6H_{2}O$
or $2Cu+6HN{O}_3\to 2Cu{\left(N{O}_3\right)}_2+NO+N{O}_2^{+}3H_{2}O$
Thus, coefficients $x$ and $y$ of $Cu$ and $HN{O}_3$ respectively are $2$ and $6$ .