At 25^{circ} C,50 g of iron reacts with HCl | vedclass

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(A) The chemical reaction is: $Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g)$.Moles of $Fe = \frac{50 \ g}{55.85 \ g \ mol^{-1}} \approx 0.89526 \

Vedclass · Accepted Answer

$2218$

Vedclass · Answer

(A) The chemical reaction is: $Fe(s) + 2HCl(aq) ightarrow FeCl_2(aq) + H_2(g)$.Moles of $Fe = \frac{50 \ g}{55.85 \ g \ mol^{-1}} \approx 0.89526 \ mol$.Since $1 \ mol$ of $Fe$ produces $1 \ mol$ of $H_2(g)$,moles of $H_2(g)$ produced $(n)$ $= 0.89526 \ mol$.Work done $(w)$ $= -P_{ext} \Delta V = -\Delta n_g RT$.Given $T = 25^{\circ} C = 298 \ K$,$R = 8.314 \ J \ mol^{-1} \ K^{-1}$,and $P_{ext} = 1 \ bar$.$w = -nRT = -0.89526 \ mol 	imes 8.314 \ J \ mol^{-1} \ K^{-1} 	imes 298 \ K$.$w = -2218.059 \ J$.The magnitude of work done by the gas is $2218 \ J$.

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