At $$NTP,$$  $$1\,L$$  of $${O_2}$$  reacts with $$3\,L$$  of carbon monoxide. What will be the volume of $$CO$$  and $$C{O_2}$$  after the reaction ?

Solution :
$$\mathop {2CO}\limits_{2\,vol} + \mathop {{O_2}}\limits_{1\,vol} \to \mathop {2C{O_2}}\limits_{2\,vol} $$
$$1\,vol$$  of $${O_2}$$  reacts with $$2\,vol$$  of $$CO$$
$$1\,L$$  of $${O_2}$$  reacts with $$2\,L$$  of $$CO$$
$$CO$$  left after reaction $$= 3 - 2 = 1 L$$
$$1\,L$$  of $${O_2}$$  produces $$2\,L$$  of $$C{O_2}.$$
Hence, after the reaction, $$CO = 1\,L,C{O_2} = 2\,L$$