Consider the following data. (i) 2Al(s) + 6HCl(aq) arrow...

The given reactions with their enthalpy changes are:
(i) $2Al(s) + 6HCl(aq) \rightarrow Al_2Cl_6(aq) + 3H_2(g)$, $\Delta H_1 = -1200$ kJ/mol
(ii) $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$, $\Delta H_2 = -164$ kJ/mol
(iii) $HCl(g) + aq \rightarrow HCl(aq)$, $\Delta H_3 = -83$ kJ/mol
(iv) $Al_2Cl_6(s) + aq \rightarrow Al_2Cl_6(aq)$, $\Delta H_4 = -663$ kJ/mol

The required reaction for the enthalpy of formation of anhydrous solid $Al_2Cl_6$ is:
$2Al(s) + 3Cl_2(g) \rightarrow Al_2Cl_6(s)$

This target equation can be obtained by applying the following algebraic operations on the given equations:
Equation (i) $+ 3 \times$ Equation (ii) $+ 6 \times$ Equation (iii) $-$ Equation (iv)

Let us verify by adding the modified equations:
$2Al(s) + 6HCl(aq) \rightarrow Al_2Cl_6(aq) + 3H_2(g)$
$3H_2(g) + 3Cl_2(g) \rightarrow 6HCl(g)$
$6HCl(g) + aq \rightarrow 6HCl(aq)$
$Al_2Cl_6(aq) \rightarrow Al_2Cl_6(s) + aq$

Summing the above reactions and canceling common terms on both sides, we get:
$2Al(s) + 3Cl_2(g) \rightarrow Al_2Cl_6(s)$

The enthalpy of formation is calculated as:
$\Delta H_f = \Delta H_1 + 3\Delta H_2 + 6\Delta H_3 - \Delta H_4$
$\Delta H_f = -1200 + 3(-164) + 6(-83) - (-663)$
$\Delta H_f = -1200 - 492 - 498 + 663$
$\Delta H_f = -2190 + 663 = -1527$ kJ/mol

Answer: $-1527$ kJ mol$^{-1}$