Respuesta :

syed514
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