Question

The volume of H2 evolved at STP when 0.9 g of Al (molar mass: 27 g mol-1) is dissolved in
excess of dilute H2SO4 is
(a) 0.58 L
(b) 1.12 L
(c) 2.40 L
(d) 2.9 L​

Answer 1

Answer:

1.12 L

Explanation:

2Al + 3H2SO4 = Al2(SO4)3 + 3H2

2 mole Al = 3 mole H2 gas

54g Al = 3 x 22.4 L

1g Al = 6.72/54

0.9g Al = 0.9 x 6.72/54

            =6.048/54

            =1.12 L

Answer 2

Answer:

Concept:

According to Avogadro’s law, the volume of one mole of any gas at Standard  Temperature and Pressure (STP = 273 K and 1 atm) is 22.4 L

Explanation:

Two important Gas Laws are required in order to convert the experimentally determined volume of hydrogen gas to that at STP.

1. Dalton's partial pressure law.

2. Combined gas law

$Mass \ of Al dissolved =0.9 \mathrm{~g} \\Equivalent weight of A l=9 g \quad\left\{\frac{27}{3}\right\} \\no. of eq. of Al dissolved =\frac{0.9}{9}=0.1 \mathrm{eq} \\no of eq of \mathrm{H}_{2} evolved =0.1 \mathrm{eq}$

$Mass \ of \ \mathrm{H}_{2} evolved 0.1 \times 1=0.1 \mathrm{~g} \\no. of moles of H_{2} evolved =\frac{0.1}{2}=\frac{1}{20} \\Molar volume at STP =22.4 \mathrm{~L} \\Volume of \mathrm{H}_{2} evolved =22.4 \times \frac{1}{20}$

                                  = 1.12 L

Hence, The volume of H2 evolved at STP when 0.9 g of Al (molar mass: 27 g mol-1) is dissolved in excess of dilute H2SO4 is 1.12L  i.e., option (c)

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