JEE MAIN - Chemistry (2023 - 31st January Evening Shift - No. 14)
Assume carbon burns according to the following equation :
$2 \mathrm{C}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{CO}(\mathrm{g})$
when $12 \mathrm{~g}$ carbon is burnt in $48 \mathrm{~g}$ of oxygen, the volume of carbon monoxide produced is ___________ $\times 10^{-1} \mathrm{~L}$ at STP [nearest integer]
[Given: Assume $\mathrm{CO}$ as ideal gas, Mass of $\mathrm{C}$ is $12 \mathrm{~g} \mathrm{~mol}^{-1}$, Mass of $\mathrm{O}$ is $16 \mathrm{~g} \mathrm{~mol}^{-1}$ and molar volume of an ideal gas at STP is $22.7 \mathrm{~L} \mathrm{~mol}^{-1}$ ]
$2 \mathrm{C}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{CO}(\mathrm{g})$
when $12 \mathrm{~g}$ carbon is burnt in $48 \mathrm{~g}$ of oxygen, the volume of carbon monoxide produced is ___________ $\times 10^{-1} \mathrm{~L}$ at STP [nearest integer]
[Given: Assume $\mathrm{CO}$ as ideal gas, Mass of $\mathrm{C}$ is $12 \mathrm{~g} \mathrm{~mol}^{-1}$, Mass of $\mathrm{O}$ is $16 \mathrm{~g} \mathrm{~mol}^{-1}$ and molar volume of an ideal gas at STP is $22.7 \mathrm{~L} \mathrm{~mol}^{-1}$ ]
Answer
227
Explanation
_31st_January_Evening_Shift_en_14_1.png)
Limiting reagent is carbon. One mole carbon produces one mole CO. Hence, volume at STP is 227 $$ \times $$ 10-1 litre.
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