Question

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

• A. 3mol
• B. 4mol
• C. 1 mol
• D. 2mol

Answer 1

Answer: (D)

Moles of $H_2=\frac{10}{2}=5\text{mol}$
Moles of $O_2=\frac{64}{32}=2\text{mol}$
$\underset{(5mol)}{2H_2}+\underset{(2mol)}{O_2}\to \underset{(4mol)}{2H_{2}O}$
Here ${\text{O}}_{\text{2}}$ is limiting reagent.
So, amount of product,
${\text{H}}_{\text{2}}\text{O}$
obtained depends on the amount of
${\text{O}}_{\text{2}}\text{.}$
According to equation, 1 mole $O_2$ on reaction with $H_2$ gives = 2 mole of $H_{2}O$ 2 mole $O_2$ on reaction with $H_2$ will give
$=\frac{2\times 2}{1}=4\text{mole}\text{of}$
Note 1 mole of $H_2$ is in excess.