If 3.365 text{ g} of ethanol (l) is burnt com | vedclass

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(B) $1$. Calculate the moles of ethanol $(C_2H_5OH)$: Molar mass = $(2 	imes 12) + (6 	imes 1) + 16 = 46 	ext{ g/mol}$. Moles = $3.365 	ext{ g} /

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$3$

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(B) $1$. Calculate the moles of ethanol $(C_2H_5OH)$: Molar mass = $(2 	imes 12) + (6 	imes 1) + 16 = 46 	ext{ g/mol}$. Moles = $3.365 	ext{ g} / 46 	ext{ g/mol} = 0.07315 	ext{ mol}$.$2$. Calculate the enthalpy of combustion $(\Delta H_{comb})$: $\Delta H_{comb} = -99.472 	ext{ kJ} / 0.07315 	ext{ mol} \approx -1360 	ext{ kJ/mol}$.$3$. Use the combustion equation: $C_2H_5OH(l) + 3O_2(g) ightarrow 2CO_2(g) + 3H_2O(l)$.$4$. $\Delta H_{comb} = [2 	imes \Delta H_f^\circ(CO_2) + 3 	imes \Delta H_f^\circ(H_2O)] - [\Delta H_f^\circ(C_2H_5OH) + 3 	imes \Delta H_f^\circ(O_2)]$.$5$. Given standard values: $\Delta H_f^\circ(CO_2) = -393.5 	ext{ kJ/mol}$, $\Delta H_f^\circ(H_2O) = -285.8 	ext{ kJ/mol}$, $\Delta H_f^\circ(O_2) = 0$.$6$. $-1360 = [2(-393.5) + 3(-285.8)] - \Delta H_f^\circ(C_2H_5OH)$.$7$. $-1360 = [-787 - 857.4] - \Delta H_f^\circ(C_2H_5OH) \Rightarrow -1360 = -1644.4 - \Delta H_f^\circ(C_2H_5OH)$.$8$. $\Delta H_f^\circ(C_2H_5OH) = -1644.4 + 1360 = -284.4 	ext{ kJ/mol}$.$9$. The magnitude $|\Delta H_f^\circ| = 284.4 	ext{ kJ/mol} = 2.844 	imes 10^2 	ext{ kJ/mol}$. Rounding to the nearest integer gives $3 	imes 10^2 	ext{ kJ/mol}$.

If $3.365 \text{ g}$ of ethanol $(l)$ is burnt completely in a bomb calorimeter at $298.15 \text{ K}$, the heat produced is $99.472 \text{ kJ}$. The $|\Delta H_f^\circ|$ of ethanol at $298.15 \text{ K}$ is . . . . . . $\times 10^2 \text{ kJ mol}^{-1}$.

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