I am stuck on this problem, and am not sure how to cancel some of the compounds like HCl.
please see attached file for question:)
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Let us mark given equations as 1,2,3,4
Top equations is desired equation
∆H of every reaction is now ∆H₁  , ∆H₂   , ∆H₃   , ∆H₄
Aim is to get the desired equation from given reactions.
To get 2Al(s) on the reactant side I will take equation 1 as such.
To get 3Cl₂ on the reactant side we will take equation 3 and multiply it by 3 also.
To get 2AlCl₃[/sub] on the product side we will reverse equation 4 and multiply it by 2.
Now to get rid of HCl we will take equation 2 and multiply it by 6.
Do same operations on ∆H
∆H(reaction) = ∆H₁ + 6 ∆H₂ + 3 ∆H₃ - 2  ∆H₄

If I understood  correctly then

ΔH = -1049+6*(-74,8)+3*(-1845)-2*(-232) =-6386,8 kJ

But Literaturen value says - 704,2 kJ/mol or for 2 mol - 1408,4 kJ

Where is the error

https://www.periodensystem-online.de/index.php?el=13&id=compound&cpid=54