Calculate the work done in Joules (J) when 50 | vedclass

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(C) The chemical reaction is: $Fe(s) + 2HCl(aq) 	o FeCl_2(aq) + H_2(g)$.Number of moles of $Fe = \frac{50 \ g}{56 \ g/mol} = 0.8928 \ mol$.Since $1 \

Vedclass · Accepted Answer

$-2212.39$

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(C) The chemical reaction is: $Fe(s) + 2HCl(aq) 	o FeCl_2(aq) + H_2(g)$.Number of moles of $Fe = \frac{50 \ g}{56 \ g/mol} = 0.8928 \ mol$.Since $1 \ mol$ of $Fe$ produces $1 \ mol$ of $H_2$,moles of $H_2$ produced = $0.8928 \ mol$.The volume of $H_2$ gas produced at $25 \ ^oC$ $(298 \ K)$ and $1 \ atm$ is calculated using the ideal gas law $PV = nRT$:$V = \frac{nRT}{P} = \frac{0.8928 \ mol 	imes 0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1} 	imes 298 \ K}{1 \ atm} = 21.842 \ L$.Work done $(W)$ in an open beaker is given by $W = -P_{ext} \Delta V$.Assuming the initial volume of the solid/liquid is negligible compared to the gas volume,$\Delta V \approx V_{gas} = 21.842 \ L$.$W = -1 \ atm 	imes 21.842 \ L = -21.842 \ L \cdot atm$.To convert $L \cdot atm$ to Joules $(J)$,use the conversion factor $1 \ L \cdot atm = 101.325 \ J$:$W = -21.842 	imes 101.325 \ J = -2213.13 \ J$.Rounding to the nearest provided option,the correct answer is $-2212.39 \ J$.

$\Delta_{\text{r}}H^0$	$\Delta_{\text{r}}S^0$	$\Delta_{\text{r}}G^0$	Description
$(a) (+)$	$(-)$	$(+)$	$(i) \text{ Non-spontaneous at all temperatures}$
$(b) (-)$	$(-)$	$(-)$	$(ii) \text{ Spontaneous at low temperatures}$
$(c) (-)$	$(+)$	$(-)$	$(iii) \text{ Spontaneous at all temperatures}$

Calculate the work done in Joules $(J)$ when $50 \ g$ of iron is dissolved in $HCl$ in an open beaker at $25 \ ^oC$. The atmospheric pressure is $1 \ atm$.

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