Question

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

• A. $0.2c{m}^3$
• B. $0.4c{m}^3$
• C. $224c{m}^3$
• D. $4480c{m}^3$

Answer 1

Answer: (A), (D)

$4480c{m}^3$
$0.2c{m}^3$

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

According to the given equation 40 gr of $Ca$ gives 22400 $c{m}^3$ of $H_2$

8 gr of $Ca$ gives X $c{m}^3$ of $H_2$

$X=\frac{8\times 22400}{40}=4480c{m}^3$

Hence option D is correct.