SO_{2}Cl_{2} on reaction with excess of water | vedclass

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(C) Let the number of moles of $SO_{2}Cl_{2}$ be $x$.According to the reaction: $SO_{2}Cl_{2} + 2H_{2}O \rightarrow H_{2}SO_{4} + 2HCl$.From the stoic

Vedclass · Accepted Answer

$4$

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(C) Let the number of moles of $SO_{2}Cl_{2}$ be $x$.According to the reaction: $SO_{2}Cl_{2} + 2H_{2}O ightarrow H_{2}SO_{4} + 2HCl$.From the stoichiometry,$1 \, mole$ of $SO_{2}Cl_{2}$ produces $1 \, mole$ of $H_{2}SO_{4}$ and $2 \, moles$ of $HCl$.Therefore,$x \, moles$ of $SO_{2}Cl_{2}$ produce $x \, moles$ of $H_{2}SO_{4}$ and $2x \, moles$ of $HCl$.Total moles of $H^{+}$ ions produced = $(2 	imes n(H_{2}SO_{4})) + n(HCl) = (2 	imes x) + 2x = 4x$.For complete neutralization,$n(H^{+}) = n(OH^{-})$.Given $n(NaOH) = 16 \, moles$,so $n(OH^{-}) = 16$.$4x = 16 \implies x = 4$.

$SO_{2}Cl_{2}$ on reaction with excess of water results in an acidic mixture: $SO_{2}Cl_{2} + 2H_{2}O \rightarrow H_{2}SO_{4} + 2HCl$. If $16 \, moles$ of $NaOH$ is required for the complete neutralization of the resultant acidic mixture,calculate the number of moles of $SO_{2}Cl_{2}$ used.

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