Question

When 0.45g of Zn is added to 50.0 mL of 0.95 M HCL solution, the solution inside the calorimeter heats up by 12 degrees Celsius. What is the molar enthalpy of the reaction (in kj/mol)?

Answer 1

Zn + 2HCl → ZnCl2 + H2

n(Zn)= 0.45/65= 0.007 moll

n(HCl)= 0.05x0.95 =0.0475 moll

n(ZnCl2)=0.007 moll

∆H(ZnCl2)=12 / 0.007 = 1714.2857 kj/mol

∆H(HCl)= -197 kj/ mol

∆Hreaction=1714.2857 –(-197)= 1911.2857 kj/mol