JEE MAIN - Chemistry (2022 - 25th July Morning Shift - No. 1)
$$\mathrm{SO}_{2} \mathrm{Cl}_{2}$$ on reaction with excess of water results into acidic mixture
$$\mathrm{SO}_{2} \mathrm{Cl}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{HCl}$$
16 moles of $$\mathrm{NaOH}$$ is required for the complete neutralisation of the resultant acidic mixture. The number of moles of $$\mathrm{SO}_{2} \mathrm{Cl}_{2}$$ used is :
16
8
4
2
Explanation
$$\mathrm{SO}_{2} \mathrm{Cl}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{HCl}$$
Moles of $$\mathrm{NaOH}$$ required for complete neutralization of resultant acidic mixture $$=16$$ moles
And 1 mole of $$\mathrm{SO}_{2} \mathrm{Cl}_{2}$$ produced 4 moles of $$\mathrm{H}^{+}$$.
$\therefore$ Moles of $$\mathrm{SO}_{2} \mathrm{Cl}_{2}$$ used will be $$=\frac{16}{4}=4$$ moles
Moles of $$\mathrm{NaOH}$$ required for complete neutralization of resultant acidic mixture $$=16$$ moles
And 1 mole of $$\mathrm{SO}_{2} \mathrm{Cl}_{2}$$ produced 4 moles of $$\mathrm{H}^{+}$$.
$\therefore$ Moles of $$\mathrm{SO}_{2} \mathrm{Cl}_{2}$$ used will be $$=\frac{16}{4}=4$$ moles
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