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Aluminum chloride can be formed from its elements: (i) 2Al(s)+3Cl2(g) ⟶ 2AlCl3(s) ΔH°= ? Use the...

Aluminum chloride can be formed from its elements: (i) 2Al(s)+3Cl2(g) ⟶ 2AlCl3(s) ΔH°= ?

Use the reactions here to determine the ΔH° for reaction(i):

(ii) HCl(g) ⟶ HCl(aq) ΔH(ii) ° =−74.8kJ

(iii) H2(g)+Cl2(g) ⟶ 2HCl(g) ΔH(iii) ° =−185kJ

(iv) AlCl3(aq) ⟶ AlCl3(s) ΔH(iv) ° =+323kJ/mol

(v) 2Al(s)+6HCl(aq) ⟶ 2AlCl3(aq)+3H2(g) ΔH(v) ° =−1049kJ

Textbook says answer is −1407 kJ I keep getting -1049 kJ - 555 kJ + 646 kJ = -958 kJ. Please help! Is there a difference when kJ/mol is involved?

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