Question

If 30 mL of $H_2$ and 20 mL of $O_2$ reacts to form water, what is left at the end of the reaction ?

• A. 10 mL of $H_2$
• B. 5 mL of $H_2$
• C. 10 mL of $O_2$
• D. 5 mL of $O_2$

Answer 1

Answer: (D)

5 mL of $O_2$

Answer 2

$\underset{2\,vol}{2H_{2}}+\underset{1vol}{O_{2}}\to 2H_{2}O$ $2$ mole of $H_{2}$ reacts with $1$vol of $O_{2}$ to give water. Therefore. $30\, mL$ of $H_{2}$ will react with $15\, mL$ of $O_{2}$ to give water. Here, $H_{2}$ is the limiting reactant. Volume of $O_{2}$ left unreacted $= 20-15 = 5\, mL$