A scientist measures the standard enthalpy change for the following reaction to be -220.4 kJ :
2H2O2(l)2H2O(l) + O2(g)
Somewhere they told you, or should've that ΔHformation for O2=0. They must've also told you that ΔHformation for H2O2= -187.6, because otherwise this problem is impossible.
Then:
ΔHrxn=ΔHf(products)-ΔHf(reactants)
ΔHrxn=(2·H2O+O2)-(2H2O2)
-220.4=(2x+0)-(2• -187.6)
-220.4=2x - -375.2
-220.4=2x+375.2
-595.6=2x
x=-297.8 kJ
answer makes sense because the actual value is -285 kJ.