Verified Solution • 2025–26 Session moderate

QJEEM21Ch55a2: JEE Main 2021 Chemistry - Class 11 JEEM21Ch55a2 Solution: At 25 C, 50 g of iron reacts with HCl to form FeCl 2 . The…

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]

Step-by-Step Verified Solution

Fe+2HCl→FeCl2+H2Fe + 2HCl \to FeC{l_2} + {H_2}

5055.85moles{{50} \over {55.85}}moles

Moles of Fe = 5055.85moles{{50} \over {55.85}}moles = Moles of H2

No. of H2 produced =5055.85moles= {{50} \over {55.85}}moles

Work done =−Pext . ΔV= - {P_{ext}}\,.\,\Delta V

=−ΔngRT= - \Delta {n_g}RT

=−5055.85×8.314×298= - {{50} \over {55.85}} \times 8.314 \times 298

= -2218.05 J

Nearest integer = 2218

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