Methane, the principal component of natural gas, is used for heating and cooking. The combustion process is CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) If 15.9 moles of CH4 react with oxygen, what is the volume of CO2 (in liters) produced at 23.7°C and 0.985 atm?

392.97 litres

Explanation:

From the equation of reaction, we can see that 1 mole of methane yielded 1 mole of carbon iv oxide. Hence, 15.9 moles of methane will yield 15.9 moles of carbon iv oxide.

At s. t. p one mole of a gas occupies a volume of 22.4L, hence 15.9 moles of a gas will occupy a volume of 22.4 * 15.9 which equals

356.16L.

Now, we can use the general gas equation to get the volume produced at the values given.

We have the following values;

V1 = 356.16L P1 = 1 atm (standard pressure) T1 = 273K (standard temperature) V2 = ? T2 = 23.7 + 273 = 296.7K P2 = 0.985 atm

The general form of the general gas equation is given as:

(P1V1) T1 = (P2V2) / T2

After substituting the values, we get V2 to be 392.97Litres