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What would be the heat released when an aqueous solution containing $0.5 mole of {HNO}_{3}$ is mixed with $0.3 mol of {OH}^{-}$ (Enthalpy of neutralization = - 57.1 kJ) ?

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$28.5 kJ$

$17.13 kJ$

$45.7 kJ$

$1.7 kJ$

answer is B.

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Detailed Solution

The enthalpy of neutralization is −57.1 kJ/mol.
0.3 mole of HNO₃ will react with 0.3 moles of OH⁻.
H⁺+ OH⁻⟶ H₂O
$∴$ Heat evolved $= 0.3 x 57.1 kJ = 17.13 kJ .$

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