Question

$(N{H}_4{)}_{2}C{r}_2{O}_7$ on heating liberates a gas. The same gas will be obtained by

• A. heating $N{H}_{4}N{O}_2$
• B. heating $N{H}_{4}N{O}_3$
• C. treating $M{g}_3{N}_2$ with $H_{2}O$
• D. heating $H_2{O}_2$ on $NaN{O}_2$

Answer 1

Answer: (A)

$(N{H}_4{)}_{2}C{r}_2{O}_7\stackrel{\text{heat}}{\to }C{r}_2{O}_3+N_2+4H_{2}O$
$N{H}_{4}N{O}_2\to N_2+2H_{2}O$
$N{H}_{4}N{O}_3\to N_{2}O+2H_{2}O$
$M{g}_3{N}_2+H_{2}O\to 3Mg(OH{)}_2+2N{H}_3$
$2NaN{O}_2+H_2{O}_2\to N{a}_2{O}_2+2HN{O}_2$
$N_2$ gas is liberated by heating $(N{H}_4{)}_{2}C{r}_2{O}_7$ and the same gas i.e., $N_2$ is liberated by heating $N{H}_{4}N{O}_2$. In all other cases $N_2$ is not a product.