Question

$10\, g$ of hydrogen and $64\, g$ of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be

• A. 2 moles
• B. 3 moles
• C. 4 moles
• D. 1 mole

Answer 1

Answer: (C)

4 moles

Answer 2

$\underset{2g}{H _{2}+1} / \underset{16g}{2 O _{2}} \rightarrow \underset{18g}{H _{2} O}$ $10\, g\, H _{2}$ required $O _{2}=80$ which is not present $64\, g\, O _{2}$ required $8\, g$ of $H _{2}$ and $H _{2}$ left $=2 g$. Thus, $O _{2}$ is the limiting reactant and $H _{2}$ is excess reactant. Hence, $H _{2} O$ formed from 64 of $O _{2}$ $=\frac{18}{16} \times 64$ $=72\, g =\frac{72}{18}\, mole$ $=4$ mole