The work done by the liberated gas when 55.85 g of iron (molar mass 55.85 g mol$^{-1}$) reacts with hydrochloric acid in an open beaker at $25^\circ$C is (a) – 2.48 kJ (b) – 2.22 kJ (c) + 2.22 kJ (d) + 2.48 kJ

The work done by the liberated gas when 55.85 g of iron (molar mass 55.85 g mol⁻¹) reacts with hydrochloric ac

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^{-1}

(a) – 2.48 kJ

(b) – 2.22 kJ

(d) + 2.48 kJ

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Step 1. $Fe(s)+2HCl(aq)\rightarrow FeCl_2(aq)+H_2(g)$ ; moles Fe $=55.85/55.85=1$ mol, so exactly 1 mole of H $_2(g)$ is liberated, giving $\Delta n_{(g)}=+1$ . …

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Q.The work done by the liberated gas when 55.85 g of iron (molar mass 55.85 g mol $^{-1}$ ) reacts with hydrochloric acid in an open beaker at $25^\circ$ C is

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Q.The work done by the liberated gas when 55.85 g of iron (molar mass 55.85 g mol−1^{-1}−1) reacts with hydrochloric acid in an open beaker at 25∘25^\circ25∘C is

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Q.The work done by the liberated gas when 55.85 g of iron (molar mass 55.85 g mol $^{-1}$ ) reacts with hydrochloric acid in an open beaker at $25^\circ$ C is

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