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Question

Heat of neutralisation for the given reaction NaOH+HClNaCl+H2O is 57.9 kJ/mole. What will be the heat released when 0.75 moles of NaOH is titrated against 1.25 moles of HCl?


A
22.56 kJ
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B
47.12 kJ
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C
43.42 kJ
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D
28.65 kJ
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Solution

The correct option is B 43.42 kJ
Given :
NaOH+HClNaCl+H2O 
Heat of neutralization for 1 mole of HCl and NaOH is 57.9 kJ/mole.
Here, NaOH is the limiting reagent and decides the extent of reaction. 
So, heat of neutralization for 0.75 mole of reactant NaOH is 0.75 mol×57.9 kJ/mole = 43.42 kJ 

Chemistry

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