If the reaction HCl 50mlmolarity 2 NaOH 50ml molarity 202 N

If the reaction HCl (50ml,molarity 2)+ NaOH (50ml, molarity 2.02) = NaCl(what molarity?) + H2O. I just need to calculate the heat formation of products, but have heat of formation for NaCl of concentration 1mol/L. How do I do it?

Solution

moles of HCL = 0.05*2 = 0.1 moles

moles of NaOH = 0.05*2.02 = 0.101 moles

so moles of NaCl formed = 0.1 moles

total volume = 0.05+0.05 = 0.1 L

so concentration of NaCl = 0.1/0.1= 1 M

dH(NaCl(aq)) = −407.27 KJ/mol

dH(HCl(aq)) = −167.2 KJ/mol

dH(H2O(l)) = –285.8 kJ/mol

dH(NaOH(aq)) = −469.15 KJ/mol

dH(rxn) = -407.27-285.8+167.2+469.15 = -56.66 KJ........ANS