Calculate the enthalpy change (or the enthalpy of reaction) for the following reaction, 2Al (s) + 3Cl2 (g) ® 2AlCl3 (s)
From the following data:
2Al (s) + 6HCl (aq) --> 2AlCl3(aq) + 3H2 (g) HCl(g) ® HCl (aq) . DH = –1049 kJ
HCl(g) --> HCl (aq) DH = –73.5 kJ
H2 (g)+Cl2 (g) -> 2HCl(g) DH=–185kJ
AlCl3 (s) --> AlCl3 (aq) DH = –323 kJ