How do I calculate the standard enthalpy change for the following reaction at 25 °C?

MgCl2 (s) + H2O (l) - - > MgO (s) + 2HCl (g)

MgCl2 (s) + H2O (l) → MgO (s) + 2 HCl (g)

Using the standard enthalpies of formation given in the source below:

(-601.24 kJ) + (2 x - 92.30 kJ) - (-641.8 kJ) - (-285.8 kJ) = + 141.76 kJ

So:

MgCl2 (s) + H2O (l) → MgO (s) + 2 HCl (g), ΔH = + 141.76 kJ