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# 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 $c{{m}^{3}}$

Last updated date: 12th Aug 2024
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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.

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.