Calculate the molar enthalpy of reaction

Question

standard enthalpy of formation below.
H20 = -285.8 kj/mole
H+ = 0.0 kj/mole
OH- = -229.9 kj/mol

H+(aq) + OH-(aq)→H2O(l)

For this, don't you do the summation of products x stoichemtry + the sum of reactants x stoich.

from there I'm not sure where to go.

Answer 1

Yes, that's what you do and when you get there there is nowhere else to go. You're subtracting kJ/mol from kJ/mol and the answer is in kJ/mol.
(1*-285.8)-(-229.9+0) = ?

Answer 2

Glycin heat at 200 degree centigrate