Calculate the number of moles of NaOH require | vedclass

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(A) The percent labeling of oleum is defined as $100 + x$, where $x$ is the mass of water required to convert all $SO_3$ in $100 \ g$ of oleum into $H

Vedclass · Accepted Answer

$2.4$

Vedclass · Answer

(A) The percent labeling of oleum is defined as $100 + x$, where $x$ is the mass of water required to convert all $SO_3$ in $100 \ g$ of oleum into $H_2SO_4$.For $118\%$ oleum, $x = 18 \ g$ of $H_2O$.The reaction is $SO_3 + H_2O \rightarrow H_2SO_4$.Since $18 \ g$ of $H_2O$ is $1 \ mole$, it reacts with $1 \ mole$ of $SO_3$ $(80 \ g)$.Thus, in $100 \ g$ of oleum, there is $80 \ g$ of $SO_3$ and $20 \ g$ of $H_2SO_4$.Moles of $SO_3 = \frac{80 \ g}{80 \ g/mol} = 1 \ mol$.Moles of $H_2SO_4 = \frac{20 \ g}{98 \ g/mol} \approx 0.204 \ mol$.Neutralization reactions are:$SO_3 + 2NaOH \rightarrow Na_2SO_4 + H_2O$ (requires $2 \ mol$ $NaOH$ per $mol$ $SO_3$)$H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$ (requires $2 \ mol$ $NaOH$ per $mol$ $H_2SO_4$)Total $NaOH$ moles $= 2(1) + 2(0.204) = 2 + 0.408 = 2.408 \ mol \approx 2.4 \ mol$.

List-$I$	List-$II$
$A$. $16 \ g \ CH_{4(g)}$	$I$. Weighs $28 \ g$
$B$. $1 \ g \ H_{2(g)}$	$II$. $60.2 \times 10^{23}$ electrons
$C$. $1 \ mole \ N_{2(g)}$	$III$. Weighs $32 \ g$
$D$. $0.5 \ mol \ SO_{2(g)}$	$IV$. Occupies $11.2 \ L$ volume at $STP$

Calculate the number of moles of $NaOH$ required to completely neutralize $100 \ g$ of $118\%$ oleum.

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