Question

Calculate the ∆Hrxn for the reaction below:

MgCl2 (aq) + 2 KOH (aq) ---> Mg(OH)2 (s) + 2 KCl (aq)

∆Hfo for MgCl2 (aq) = -797.1 kJ/mol

∆Hfo for KOH (aq) =-424.2 kJ/mol

∆Hfo for Mg(OH)2 (s) = -924.7 kJ/mol

∆Hfo for KCl (aq) = -419.6 kJ/mol

∆Hrxno = ? kJ

Don't forget the sign and write your 4 significant figures.

Answer 1

MgCl2 (aq) + 2 KOH (aq) ---> Mg(OH)2 (s) + 2 KCl (aq)

∆Hfo for MgCl2 (aq) = -797.1 kJ/mol

∆Hfo for KOH (aq) =-424.2 kJ/mol

∆Hfo for Mg(OH)2 (s) = -924.7 kJ/mol

∆Hfo for KCl (aq) = -419.6 kJ/mol

MgCl2 (aq) + 2 KOH (aq) ---> Mg(OH)2 (s) + 2 KCl (aq)

∆Hrxno = ∆Hf products - ∆Hf reactants

             = (2*-419.6-924.7) -(-797.1+2*-424.2)

            = -118.4KJ >>>>>answer