Question

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

The atomic mass of calcium is 40 g/mol. Here one mole of calcium reacts with two mole of water to form one mole of calcium hydroxide and one mole of hydrogen which is 22.4 L at STP.

Complete step by step answer:
Given,
Mass of calcium is 8 g.
Calcium reacts with water to give calcium hydroxide and releases hydrogen gas.
The given reaction between calcium and water is shown below.
$Ca + 2{H_2}O \to Ca{(OH)_2} + {H_2} \uparrow$
In this reaction, one mole of calcium reacts with 2 mole of water to form one mole of calcium hydroxide and one mole of hydrogen.
The amount of moles of any substance present can be determined by its molar mass.
The atomic mass of calcium is 40 g/mol.
So, 1 mol calcium will be equal to 40 g.
40 g of calcium gives 22.4 L of ${H_2}$ in STP. 22.4 L = 22400 $c{m^3}$.
So, 8 g of calcium will give $\Rightarrow \dfrac{{8 \times 22400}}{{40}} = 4480c{m^3}$.
Thus, the 8 g of calcium completely reacts with water to form 4480 $c{m^3}$ volume of hydrogen at STP.

So, the correct answer is Option D.

Note: Don’t get confused by the statement given that the question contains multiple correct options. Make sure to convert the volume in Litre into cubic centimeter. 1 L = 1000 mL, 1 mL = 1$c{m^3}$, 1 L = 1000 $c{m^3}$.