Use the standard enthalpies of formation for the reactants and products to solve for the δhrxn for the following reaction. (the δhf of ca(oh)2 is -986.09 kj/mol and liquid h2o is -285.8 kj/mol.) ca(s) + 2h2o(l) > ca(oh)2 (s) + h2(g)

1. Remember (sum of products) - (sum of reactants)
So ΔHrxn = 2 ΔHf [H2(g)] + ΔHf [Ca(OH)2(s)] - 2 ΔHf [H2O(l)] - ΔHf [Ca(s)]
= 2*0 + -986.09 kJ/mol - 2*(-285.8 kJ/mol) - 0

Do the math and you'll have the answer. BTW the ΔHf [H2(g)] and ΔHf [Ca(s)] were 0 because these are elements in their standard states.

HOPE THIS HELPS ;)