Question

An electric discharge is passed through a mixture containing 50 cc of $O_2$ and 50 cc of $H_2$. The volume of the gas formed (i) at room temperature (ii) at ${110}^{\circ }C$ will be:

(All measurements are made at same pressure and temperature)

• A. (i) 25 cc (ii) 50 cc
• B. (i) 50 cc (ii) 75 cc
• C. (i) 25 cc (ii) 75 cc
• D. (i) 75 cc (ii) 75 cc

Answer 1

Answer: (A)

(i) 25 cc (ii) 50 cc

$2H_2+O_2\rightleftharpoons H_{2}O\left(l\right)$

At room as $25cc$ of $O_2$ reacts with $50cc$ of $H_2$ so only $25ccof{O}_2$ is remaining.

At ${110}^{\circ }C$, As $25cc$ of $O_2$ reacts with $50cc$ of $H_{2}25cc$ of vapour is produced.

So a total of 50 cc is produced