The weight of hydrogen gas obtained from $42 g$ of $C a H_{2}$ by treatment with water is ..... gm.

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Solution

$C a H_{2} + 2 H_{2} O \to C a (O H)_{2} + 2 H_{2} ↑$
The molar mass of $C a H_{2} = 40 + 2 = 42$ g/mol.
The number of moles of $C a H_{2} = \frac{42 g}{42 g / m o l} = 1 m o l$
1 mole of $C a H_{2}$ will liberate 2 moles of $H_{2}$
The molar mass of $H_{2} = 2$ g/mol
The mass of $H_{2}$ gas obtained $= 2 g / m o l \times 1 m o l = 2 g$

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The weight of hydrogen gas obtained from 42 g of CaH2 by treatment with water is ..... gm.

CaH2+2H2O→Ca(OH)2+2H2↑The molar mass of CaH2=40+2=42 g/mol.The number of moles of CaH2=42g42g/mol=1mol1 mole of CaH2 will liberate 2 moles of H2The molar mass of H2=2 g/molThe mass of H2 gas obtained =2g/mol×1mol=2g

The weight of hydrogen gas obtained from $42 g$ of $C a H_{2}$ by treatment with water is ..... gm.