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
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