Question

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?

Answer 1

Answer:

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