Heat of neutralisation for the given reaction NaOH+HCl→NaCl+H2O is 57.9kJ/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
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
47.12 kJ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
43.42 kJ
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
28.65 kJ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C43.42 kJ Given : NaOH+HCl→NaCl+H2O Heat of neutralization for 1 mole of HCl and NaOH is 57.9kJ/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 0.75mol×57.9 kJ/mole = 43.42 kJ