The work done in the open vessel at 300text{ K} when 112text{ g} of iron reacts with dilute text{HCl} is: (text{Fe} = 56text{ g mol}^{-1}, R = 0.0821…

Reaction: $\text{Fe(s)} + 2\text{HCl(aq)} \rightarrow \text{FeCl}_2\text{(aq)} + \text{H}_2\text{(g)}$\nMoles of $\text{Fe} = 112 / 56 = 2\text{ mol}$.\nMoles of $\text{H}_2$ gas produced $= 2\text{ mol}$.\nWork done $w = -\Delta n_g RT = -2 \times (2\text{ cal K}^{-1}\text{mol}^{-1}) \times 300\text{ K} = -1200\text{ cal}$. (Using $R \approx 2\text{ cal K}^{-1}\text{mol}^{-1}$). Wait, let's recalculate accurately: $w = -P\Delta V = -\Delta n_g RT = -2 \times 2 \times 300 = -1200\text{ cal}$. Let's verify options. If $R = 2\text{ cal}$, $2 \times 2 \times 300 = 1200$. Let's check typical values: $2\text{ mol Fe} \rightarrow 2\text{ mol H}_2$, so $w = -2 \times 2 \times 300 = -1200\text{ cal}$. Wait, if the option is -2400, let's re-verify $\Delta n_g$. $\text{Fe} + 2\text{H}^+ \rightarrow \text{Fe}^{2+} + \text{H}_2$. Moles of Fe = $112/56 = 2$. So $\Delta n_g = 2$. $w = -2 \times 2 \times 300 = -1200\text{ cal}$. Let's change option A or B to match -1200 cal.