Calculate the enthalpy of formation of anhydrous aluminium chloride, `Al_(2)Cl_(6)` from the following data :
(i) `2Al(s) + 6HCl (aq) to Al_(2)Cl_(6) (aq) + 3H_(2)(g), Delta_(r)H^(@) = -1004.0 kJ`
(ii) `H_(2)(g) + Cl_(2)(aq) to 2HCl(g) , Delta_(r)H^(@) = -183.9 kJ , Delta_(r)H^(@) = -183.9 kJ`
(iii) `HCl(g) + aq to HCl(aq), Delta_(r)H^(@) = -73.2 kJ`
`Al_(2)Cl_(6)(s) + aq to Al_(2) Cl_(2)(aq) , Delta_(r)H^(@) = -643.0 kJ`

Explanation

To calculate the enthalpy of formation of anhydrous aluminium chloride, \( \text{Al}_2\text{Cl}_6 \), we will manipulate the given reactions and their enthalpy changes according to Hess's law. Let's break down the steps:

### Step 1: Write the target reaction
The target reaction for the formation of anhydrous aluminium chloride from its elements is:
\[ 2 \text{Al}(s) + 3 \text{Cl}_2(g) \rightarrow \text{Al}_2\text{Cl}_6(s) \]

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