5.6 L of methane was mixed with 60 L of air. The mixture was blown up, the reaction products were cooled.

5.6 L of methane was mixed with 60 L of air. The mixture was blown up, the reaction products were cooled. Determine the composition of the gas mixture after the reaction.

Given:
V (CH4) = 5.6 L
V (air) = 60 l

To find:
φ (gases) -?

Decision:
1) Write the reaction equation:
CH4 + 2O2 => CO2 + 2H2O;

2) Find the amount of substance CH4:
n (CH4) = V (CH4) / Vm = 5.6 / 22.4 = 0.25 mol;

3) Find the volume of N2, O2, inert gases and CO2 in the given volume of air:
V (N2) = φ (N2) * V (air) / 100% = 78.09% * 60/100% = 46.85 l;
V (O2) = φ (O2) * V (air) / 100% = 20.95% * 60/100% = 12.57 l;
V (in.gas) = ​​φ (in.gas) * V (air) / 100% = 0.94% * 60/100% = 0.56 l;
V (CO2) = V (air) – V (N2) – V (O2) – V (in.gas) = ​​60 – 46.85 – 12.57 – 0.56 = 0.02 l;

4) Find the amount of substance O2:
n (O2) = V (O2) / Vm = 12.57 / 22.4 = 0.56 mol;

5) Compare the amount of substance CH4 and O2:
The amount of CH4 is in short supply, therefore, calculations will be carried out for it;

6) Find the amount of substance O2 that reacted with CH4 (taking into account the reaction equation):
n1 (O2) = n (CH4) * 2 = 0.25 * 2 = 0.5 mol;

7) Find the amount of substance O2 that remains:
n2 (O2) = n (O2) – n1 (O2) = 0.56 – 0.5 = 0.06 mol;

8) Find the amount of O2 that remains:
V2 (O2) = n2 (O2) * Vm = 0.06 * 22.4 = 1.34 L;

9) Find the amount of substance CO2 that was formed (taking into account the reaction equation):
n1 (CO2) = n (CH4) = 0.25 mol;

10) Find the volume of CO2 generated:
V1 (CO2) = n1 (CO2) * Vm = 0.25 * 22.4 = 5.6 l;

11) Find the total CO2 volume:
V2 (CO2) = V (CO2) + V1 (CO2) = 0.02 + 5.6 = 5.62 L;

12) Find the volume of the resulting mixture:
V mixture = V (N2) + V2 (O2) + V (in.gas) + V2 (CO2) = 46.85 + 1.34 + 0.56 + 5.62 = 54.37 l;

13) Find the volume fractions of N2, O2, inert gases, CO2 in the resulting mixture:
φ (N2) = V (N2) * 100% / V mixture = 46.85 * 100% / 54.37 = 86.2%;
φ (O2) = V (O2) * 100% / V mixture = 1.34 * 100% / 54.37 = 2.5%;
φ (in.gas) = ​​V (in.gas) * 100% / V mixture = 0.56 * 100% / 54.37 = 1%;
φ (gas) = ​​100% – φ (N2) – φ (O2) – φ (in.gas) = ​​100% – 86.2% – 2.5% – 1% = 10.3%.

Answer: The volume fractions of N2, O2, inert gases, CO2 are 86.2%; 2.5%; one%; 10.3% respectively.