Find the volume of air required for complete combustion of 4.48L of carbon disulfide (CS2).

Given:
4.48L carbon disulfide (CS2), air
To find:
Find the volume of air required for complete combustion of 4.48L of carbon disulfide (CS2) -?
Decision:
Let’s compose the reaction equation:
CS2 + 3O2 = CO2 + 2SO2
1) n (CS2) = 4.48 / 22.4 = 0.2 (mol)
According to the reaction, 1 (mol) of carbon disulfide accounts for 3 (mol) oxygen, therefore:
2 n (O2) = 0.2 (mol) * 3 = 0.6 (mol)
3) V (O2) = 22.4 * 0.2 = 4.48 (l) – volume of 21% air, hence the volume of 100% air:
4) V (air) = 4.48 * 100% / 21 = 21.33 (l).
Answer: 21.33 liters – 100% air volume.