Answer to Question #130185 in General Chemistry for A

Question

A neutralization experiment using NaOH and HCl produced 986 J. The limiting reactant was NaOH, which was present in 1.94×10-2 moles.

Calculate the enthalpy change of neutralization (in kJ/mol).

Give your answer to three significant figures.

Answer 1

$NaOH+HCl \to NaCl +H_2O$

This is the neutralisation reaction.

So, the enthalply change to be calculated will be on enthalpy change per mole of $NaOH$ reacted( As it is limiting reagent)

Net Enthalply change given $(H)=986 \ J$

Moles reacted $(N) =1.94\times 10^{-2}$

Heat of reaction $(H_r) =\frac{H}{N}=\frac{986}{1.94\times 10^{-2}}J=50824.74J=50.824\ KJ$$=50.8 \ KJ$ (upto 3 significant figures)