I'm not sure if this is what you're looking for, but I suppose you would have to have the solute and the solution to balance the reaction.
P1V1 = P2V2
1.2ATM*(.7 L) = V2 * .95L
V2 = 0.884 Liters or 884.21 ml
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
The property to be used is sieving. you get a bowl and a sieve when you pour the sand and salt,the salt being the small particle sieve to the bowl while the sand remain on the sieve as residue
