Calculate the amount of energy released when 500 cm³ of methane gas at STP reacts with excess air according to the equation: $$\ce{CH4(g) + 2O2(g) -&gt; CO2(g) + 2H2O(l)} \qquad \Delta H = -890\ \mathrm{kJ\ mol^{−1}}$$

I am not totally sure how to work this out. 

What I did, though, was:
$\ce{CH4}$ $M = 16\ \mathrm{g/mol}$
So, $500\ \mathrm{cm^3} = 31.25\ \mathrm{mol}$
$\ce{CH4}$ $H_\mathrm f = -890\ \mathrm{kJ/mol}$
-> For $31.25\ \mathrm{mol}$ we have $27\,812.5\ \mathrm{kJ}$

If I am wrong, can someone explain me why?
How you find the moles from volume? Did you assume $500\ \mathrm{cm^{3}}$ is $500\ \mathrm g$? This is wrong, because the density of $\ce{CH4}$ is not $1\ \mathrm{g/cm^{3}}$. You have to use another way to calculate moles of the gas. Think about ideal gas law or how much volume for STP gas.