Question

The reaction of calcium with water is represented by the equation:
$Ca+2{{H}_{2}}O\xrightarrow{{}}Ca{{(OH)}_{2}}+{{H}_{2}}\uparrow$
What volume of ${{H}_{2}}$ at STP would be liberated when 8 $g$ of calcium completely reacts with water?
A.0.2 $c{{m}^{3}}$
B.0.4 $c{{m}^{3}}$
C.224 $c{{m}^{3}}$
D.4480 $c{{m}^{3}}$

Answer 1

Hint: The atomic weight of calcium is 40 $g$. One mole of calcium (40 $g$) reacts with water and forms one mole of hydrogen gas means 22.4 L of hydrogen gas.

Complete step by step answer:
The given balanced chemical equation is as follows
$Ca+2{{H}_{2}}O\xrightarrow{{}}Ca{{(OH)}_{2}}+{{H}_{2}}\uparrow$
Here 1 mole of calcium completely reacts with 2 moles of water and forms one mole of calcium hydroxide and one mole of hydrogen as gas.
Now the question is if 8 g of calcium reacts with water, how much hydrogen gas is going to be released.
$Ca+2{{H}_{2}}O\xrightarrow{{}}Ca{{(OH)}_{2}}+{{H}_{2}}\uparrow$
40 $g$ 22.4 L
22400 ml
22400 $c{{m}^{3}}$
Means 40 g of calcium reacts with water and forms 22.4 L of hydrogen.
We can write 22.4 L of hydrogen in to 22400 $ml$or 22400 $c{{m}^{3}}$.
8 $g$ of calcium will produce hydrogen gas $\begin{aligned} & =\dfrac{22400}{40}\times 8\text{ }c{{m}^{3}} \\\ & =44800\text{ }c{{m}^{3}} \\\ \end{aligned}$

So, 8 $g$ of calcium reacts with water and produces 4480 $c{{m}^{3}}$of hydrogen gas.
So, the correct option is D.

Note: Don’t be confused with the symbols L, ml , $c{{m}^{3}}$.
We know that one liter (L) = 1000 ml
One milliliter (ml) = 1 $c{{m}^{3}}$
Finally one liter (L) = 1000 $c{{m}^{3}}$ for gases.
One liter (L) = 1000 ml = 1000 $c{{m}^{3}}$in case of gases.