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The reaction of calcium with water is represented by the equation \[Ca+2{{H}_{2}}O\to Ca{{(OH)}_{2}}+{{H}_{2}}\]
What volume of \[{{H}_{2}}\] at STP (in \[c{{m}^{3}}\]) would be liberated when 8 g of calcium completely reacts with water?
(A) 0.2
(B) 0.4
(C) 2240
(D) 4480

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Last updated date: 26th Feb 2024
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IVSAT 2024
Answer
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Hint: In order to solve this question, first, check the stoichiometric coefficients. Then, apply Avogadro’s law to find the relationship between moles and volume. One mole of any gas at STP occupies 22.4 L.

Complete step by step answer:
\[Ca+2{{H}_{2}}O\to Ca{{(OH)}_{2}}+{{H}_{2}}\]
From the chemical reaction given above, we can see that 1 mole of Calcium reacts with 2 moles of water to give 1 mole of calcium hydroxide and water each.
According to the question, the Calcium in the reactant side reacts completely. So, Calcium is a limiting reagent.
According to Avogadro’s hypothesis, 1 mol = 22.4 L
Since 1 L = 1000 \[c{{m}^{3}}\]
Therefore, 22.4 L = 22400 \[c{{m}^{3}}\]
Also, molecular weight of Calcium = 40 g
Since, 1 mole of Calcium gives 1 mole of \[{{H}_{2}}\] gas, we can say that 40 g of Calcium will give 22400 \[c{{m}^{3}}\]of \[{{H}_{2}}\].
So, 1 g of Calcium will give \[\dfrac{22400}{40}c{{m}^{3}}\] \[{{H}_{2}}\].
Hence, 8 g of Calcium will give \[\dfrac{\text{22400}}{\text{40}}\text{ x 8 c}{{\text{m}}^{\text{3}}}\text{ }{{\text{H}}_{\text{2}}}\] = \[\text{4480 c}{{\text{m}}^{\text{3}}}\text{ }{{\text{H}}_{\text{2}}}\]
Therefore, the answer is – option (d).

Additional Information: Ideal gas equation relates pressure (P), Volume (V), Number of moles (n), Gas Constant (R) and temperature (T) by the equation – PV = nRT.

Note: Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules. For a given mass of an ideal gas, the volume and amount (moles) of the gas are directly proportional if the temperature and pressure are constant”.