JEE MAIN - Chemistry (2021 - 16th March Evening Shift - No. 17)

At 25$$^\circ$$C, 50 g of iron reacts with HCl to form FeCl2. The evolved hydrogen gas expands against a constant pressure of 1 bar. The work done by the gas during this expansion is _________ J. (Round off to the Nearest Integer).

[Given : R = 8.314 J mol$$-$$1 K$$-$$1. Assume, hydrogen is an ideal gas] [Atomic mass of Fe is 55.85 u]
Answer
2218

Explanation

$$Fe + 2HCl \to FeC{l_2} + {H_2}$$

$${{50} \over {55.85}}moles$$

Moles of Fe = $${{50} \over {55.85}}moles$$ = Moles of H2

No. of H2 produced $$ = {{50} \over {55.85}}moles$$

Work done $$ = - {P_{ext}}\,.\,\Delta V$$

$$ = - \Delta {n_g}RT$$

$$ = - {{50} \over {55.85}} \times 8.314 \times 298$$

= -2218.05 J

Nearest integer = 2218

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