Using this information together with the standard enthalpies of formation of o2(g), co2(g), and h2o(l) from appendix c, calculate the standard enthalpy of formation of acetone.
complete combustion of 1 mol of acetone (c3h6o) liberates 1790 kj:
c3h6o(l)+4o2(g)? 3co2(g)+3h2o(l)? h? =? 1790kj

ΔHacetone = - 247.5 kJ/mol

Explanation:

The enthalpy equation is as follows

ΣnΔHproducts – ΣmΔHreactants =ΔHreaction

3×ΣnΔH(CO2(g)) + 3×ΣnΔH(H2O) - ΣnΔH (C3H6O(l)) =ΔHreaction = -1790 kJ/mol

[3(-393.5) + 3(-285.8)] – ΔHacetone

= -1790 kJ/mol

(-1180.5 – 857.4)kJ/mol - ΔHacetone =

-1790 kJ/mol

-2037.9 kJ/mol - ΔHacetone

= -1790kJ/mol

-2037.9 kJ/mol + 1790kJ/mol = ΔHacetone

- 247.5 kJ/mol = ΔHacetone

ΔHacetone = - 247.5 kJ/mol

dynamic equilibrium is the answer.

if you're asking what comes out, it's steam.