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 \[\begin{align}
  & =\dfrac{22400}{40}\times 8\text{ }c{{m}^{3}} \\
 & =44800\text{ }c{{m}^{3}} \\
\end{align}\]

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.