Question

The reaction of calcium with water is represented by the equation $Ca+2H_{2}O\to Ca(OH{)}_2+H_2$
What volume of $H_2$ at STP would be liberated when $8gm$ of calcium completely reacts with water?

• A. $0.2c{m}^3$
• B. $0.4c{m}^3$
• C. $2240c{m}^3$
• D. $4480c{m}^3$

Answer 1

The correct option is D $4480c{m}^3$

Here, calculation is done, using the balanced chemical equation as follows:

$Ca+2H_{2}O\to Ca(OH{)}_2+H_2$

According to the stoichiometry of reaction, $40gm$ of $Ca$ on complete reaction with water liberates $=2gm{H}_2$

$\therefore 8gm$ of $Ca$, on complete reaction with water will liberates $=\frac{2}{40}\times 8gm{H}_2$

$=0.40gm{H}_2$

$=\frac{0.40}{2}\times 22400c{m}^3$

$=4480c{m}^3$ of $H_2$ at S.T.P.