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,
$xCu + yHNO _{3} \longrightarrow Cu \left( NO _{3}\right)_{2}+ NO$ $+ NO _{2}+ H _{2} O$
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 $N O$ and $N O_{2}$ respectively are
$3 Cu +8 HNO _{3} \rightarrow 3 Cu \left( NO _{3}\right)_{2}+2 NO +4 H _{2} O \ldots$ (i)
$Cu +4 HNO _{3} \rightarrow Cu \left( NO _{3}\right)_{2}+2 NO _{2}+2 H _{2} O \ldots$ (ii)
Here $N O$ and $N O_{2}$ are evolved in equal volumes, hence, on adding Eqs. (i) and (ii)
$4 Cu +12 HNO _{3} \rightarrow 4 Cu \left( NO _{3}\right)_{2}+2 NO +2 NO _{2}+6 H _{2} O$
or $2 Cu +6 HNO _{3} \rightarrow 2 Cu \left( NO _{3}\right)_{2}+ NO + NO _{2}^{+} 3 H _{2} O$
Thus, coefficients $x$ and $y$ of $C u$ and $H N O_{3}$ respectively are $2$ and $6$ .