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(C) The balanced chemical equation for the neutralization reaction is: $2 NaOH + H_2SO_4 \longrightarrow Na_2SO_4 + 2 H_2O$.Let the volume of $2 \ M$

Vedclass · Accepted Answer

$84$

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(C) The balanced chemical equation for the neutralization reaction is: $2 NaOH + H_2SO_4 \longrightarrow Na_2SO_4 + 2 H_2O$.Let the volume of $2 \ M$ $NaOH$ be $V \ L$ and the volume of $3 \ M$ $H_2SO_4$ be $(4 - V) \ L$.For complete neutralization,the moles of $H^+$ must equal the moles of $OH^-$.Moles of $OH^- = 2 	imes V = 2V$.Moles of $H^+ = 2 	imes (3 	imes (4 - V)) = 6(4 - V)$.Equating the moles: $2V = 6(4 - V)$ $\Rightarrow 2V = 24 - 6V$ $\Rightarrow 8V = 24$ $\Rightarrow V = 3 \ L$.Thus,moles of $NaOH = 2 	imes 3 = 6 \ mol$.Since $1 \ mol$ of $OH^-$ reacts with $1 \ mol$ of $H^+$ to release $14 \ kcal$ of heat,$6 \ mol$ of $OH^-$ will release $6 	imes 14 = 84 \ kcal$ of heat.

Calculate the maximum possible heat which can be released if a $4 \ L$ aqueous solution is to be formed by mixing appropriate volumes of $2 \ M$ $NaOH$ and $3 \ M$ $H_2SO_4$ solution. Given $\Delta H_{\text{neutralisation, SA/SB}} = -14 \ kcal/mol$.

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