How do you calculate the standard enthalpy change for the reaction at 25 ∘ C. ??

Question

Mg ( OH ) 2 ( s ) + 2 HCl ( g ) ⟶ MgCl 2 ( s ) + 2 H 2 O ( g )

Answer 1

Mg(OH)₂( s ) + 2HCl ( g ) ==> MgCl₂( s ) + 2H₂O(g)

Look up the ∆H_fº of reactants and products. Then ∑produts - ∑reactants = ∆H_reaction

These are the values I found:

Mg(OH)₂(s) = -924.7 kJ/mol

HCl(g) = -92.5 kJ/mol

MgCl₂(s) = -641.8 kJ/mol

H₂O(g) = -241.8 kJ/mol

(-641.8 + 2x-241.8) - (-924.7 + 2x-92.5) = ∆H_reaction

-1125.4 - (-1109.7) = ∆H_reaction

∆H_reaction = -15.7 kJ