Question

Suppose you are investigating the reaction: M(s) + 2 HCl(aq) → MCl2(aq) + H2(g). You weigh out a 0.250 gram piece of metal and combine it with 72.9 mL of 1.00 M HCl in a coffee-cup calorimeter. If the molar mass of the metal is 42.29 g/mol, and you measure that the reaction absorbed 155 J of heat, what is the enthalpy of this reaction in kJ per mole of limiting reactant? Enter your answer numerically to three significant figures in units of kJ/mol.

Answer 1

M(s) + 2 HCl(aq) → MCl2(aq) + H2(g)

no of moles of HCl   = molarity * volume in L

                               = 1*0.0729   = 0.0729 moles

no of moles of M     = W/G.A. Wt

                                = 0.25/42.29   = 0.0059 moles

M(s) + 2 HCl(aq) → MCl2(aq) + H2(g)

2 moles of HCl react with 1 mole of M

0.0729moles of HCl react with = 1*0.0729/2 = 0.03645 moles of M is required

M is limiting reactant

H   = 155J

          = 155J/0.0059mole

          = 26271J/mole

          = 26.271KJ/mole round off

           = 26.3KJ/mole >>>>answer