Question

The following reaction occurs at STP.

$2H_{2}O\left(I\right)-2H_2\left(g\right)+O_2\left(g\right)$.
How many liters of hydrogen gas can be produced by the breakdown of 72 grams of water?

• A. 5.6 liters
• B. 11.2 liters
• C. 22.4 liters
• D. 44.8 liters
• E. 89.6 liters

Answer 1

Answer: (E)

89.6 liters

$2H_{2}O\to 2H_2+O_2$
As per reaction, 2 moles of $H_{2}O(18\times 2=36g)$ produces 2 moles of $H_2(2\times 22.4=44.8L)$.
Since, 1 mol of gas has 22.4 L at STP.
Moles of water present in 72 g = $\frac{72}{18}=4moles$
As 2 moles of $H_{2}O$ produces 44.8 L of $H_2$ gas, so 4 moles of $H_{2}O$ will produces $=44.8\times 2=89.6L$ of $H_2$ gas.