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The heat of Combustion of acetaldehyde to carbon dioxide and water is $-1172 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$. Calculate amount of heat liberated when $66 \mathrm{~g}$ of acetaldehyde were completely oxidised ? (at. mass $\mathrm{C}=12, \mathrm{H}=1,0=16$ )
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The correct answer is:
$1758 \mathrm{~kJ}$
$\mathrm{CH}_{3} \mathrm{CHO}+2.5 \mathrm{O}_{2} \longrightarrow 2 \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \quad \Delta \mathrm{H}=-1172 \mathrm{~kJ} \mathrm{~mol}^{-1}$
For $44 \mathrm{~g}$ of $\mathrm{CH}_{3} \mathrm{CHO}, \Delta \mathrm{H}=-1172 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\therefore 66 \mathrm{~g}$ of $\mathrm{CH}_{3} \mathrm{CHO}, \Delta \mathrm{H}=\frac{-1172 \times 66}{44}=-1758 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\therefore$ The amount of heat liberated when $66 \mathrm{~g}$ of acetaldehyde were completely oxidised is $1758 \mathrm{~kJ} \mathrm{~mol}^{-1}$
For $44 \mathrm{~g}$ of $\mathrm{CH}_{3} \mathrm{CHO}, \Delta \mathrm{H}=-1172 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\therefore 66 \mathrm{~g}$ of $\mathrm{CH}_{3} \mathrm{CHO}, \Delta \mathrm{H}=\frac{-1172 \times 66}{44}=-1758 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\therefore$ The amount of heat liberated when $66 \mathrm{~g}$ of acetaldehyde were completely oxidised is $1758 \mathrm{~kJ} \mathrm{~mol}^{-1}$
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