MCQ
One mol of methanol when burnt in excess of $O_2$ gives out $723 \,kJ\, mol^{-1}$ Heat. If $1\, mol$ of $O_2$ is used, what will be the amount of heat evolved......$kJ$
- A$723$
- B$924$
- ✓$482$
- D$241$
✓
Answer
Correct option: C.
$482$
c
$\mathrm{CH}_{3} \mathrm{OH}_{(l)}+\frac{3}{2} \mathrm{O}_{2(\mathrm{g})} \rightarrow \mathrm{CO}_{2(\mathrm{g})}+2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}$
$\mathrm{CH}_{3} \mathrm{OH}_{(l)}+\frac{3}{2} \mathrm{O}_{2(\mathrm{g})} \rightarrow \mathrm{CO}_{2(\mathrm{g})}+2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}$
$\frac{3}{2} \,\mathrm{mol} \,\mathrm{O}_{2} \rightarrow 723\, \mathrm{KJ}$ heat is evolved
$1\,\mathrm{mol}\, \mathrm{O}_{2} \rightarrow \frac{723}{3 / 2}=482 \,\mathrm{KJ}$ heat will be evolved
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