Calculate the maximum possible heat which can | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$84$

Vedclass · Answer

(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 $\Delta H_{\text{neutralisation, SA/SB}} = -14 \ kcal/mol$.

Similar Questions

Moles of conc. $HNO_3$ required to oxidise one mole of sulphur is :
$S + 6HNO_3 \rightarrow H_2SO_4 + 6NO_2 + 2H_2O$

The mass of ammonia in grams produced when $2.8 \ kg$ of dinitrogen quantitatively reacts with $1 \ kg$ of dihydrogen is.............

The heat of neutralisation of a strong acid and a strong base is $13.7 \ kcal$. The heat released when $0.6 \ mole$ $HCl$ solution is added to $0.25 \ mole$ of $NaOH$ is: (in $kcal$)

If the heat of combustion of methane is $-100 \ kJ/mol$,then calculate the heat evolved when $32 \ g$ of methane reacts with $32 \ g$ of oxygen. (in $kJ$)

The amount of $BaSO_4$ formed upon mixing $100 \ mL$ of $20.8 \%$ $BaCl_2$ solution with $50 \ mL$ of $9.8 \%$ $H_2SO_4$ solution will be ............. $g$
( $Ba = 137, Cl = 35.5, S = 32, H = 1$ and $O = 16$ )

Vedclass Test Series

Exam Paper Generator

Online Exam Module