The amount of heat evolved when 500 cm3 of 0.1 M HCl is mixed with 200 cm3 of 0.2 M NaoH is (A) 2.292 kJ (B) 1.292 kJ (C) 22.9 kJ (D) 0.292 kJ

(A)No. Of moles of H ion=0.1×500/1000=0.05(N=M×V/1000 IN L)No. Of moles of OH ion=0.2×200/1000=0.04As OH is the limiting reagent.Neutralisation of 1 mole of OH provide=57.1 " " 0.04 " " " " =57.1×0.04 =2.292kj

Dear Student,
Please find below the solution to your problem.

500cm3 of 0.1 M HCl corresponds to 0.5×0.1=0.05 moles of hydrogen ions.
200cm3 of 0.2 M NaOH corresponds to 0.2×0.2=0.04 moles of hydroxide ions.
sodium hydroxide is completely neutralize by Hcl
The enthalpy of neutralisation is 57.3 kJ/mol.
For 0.04 moles, the enthalpy of neutralisation is 0.04×57.3=2.292kJ

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