Question

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

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

Answer 1

The correct option is C

(i) 25 c.c. (ii) 75 c.c

$2H_2+O_2\to 2H_{2}O$

$2H_{2\left(g\right)}+O_{2\left(g\right)}\to 2H_2{O}_{\left(l\right)}$ at room temp

$2H_{2\left(g\right)}+O_{2\left(g\right)}\to 2H_2{O}_{\left(g\right)}$ at ${110}^{\circ }C$

At room temperature 25mL of $O_2$ gas would be left unreacted while at ${110}^{\circ }C$ 50mL of$H_{2}O$ and 25mL of$O_2$ will be present.