Carbon monoxide burns : #sf(2CO+O_2rarr2CO_2)#. If you have 1 litre of #sf(CO)# how much #sf(CO_2)# is formed and how much #sf(O_2)# is required ?

In order for all the carbon monoxide to react, you'd need 0.50 L of oxygen gas. Under these conditions, the reaction will produce 1.0 L of carbon dioxide.

Start with the balanced chemical equation for your reaction

#color(red)(2)CO_((g)) + O_(2(g)) -> 2CO_(2(g))#

Notice that you have a #color(red)(2):1# mole ratio between carbon monoxide and oxygen gas. This means that, regardless of how many moles of carbon dioxide react, you'll always need 2 times less moles of oxygen in order for the reaction to take place.

Likewise, the #1:1# mole ratio that exists between carbon monoxide and carbon dioxide means that you'll produce the same number of moles of the latter as you have of the former.

When it comes to gases, you'll find that moles and volume are proportional to one another if temperature and pressure are constant - this is know as Avogadro's Law.

This means that the mole ratios become volume ratios. So, if you want all the carbon monoxide to react, you'll need

#1.0cancel("L "CO) * ("1 L "O_2)/(color(red)(2)cancel("L "CO)) = color(green)("0.50 L")# #O_2#

Likewise, you'll get

#1.0cancel("L "CO) * ("2 L "CO_2)/(2cancel("L" CO)) = color(green)("1.0 L")# #CO#

#0.5"litre"# of oxygen is required and #1"litre"# of carbon dioxide is formed.

#2CO_((g))+O_(2(g))rarr2CO_(2(g))#

So:

2mol #CO# reacts with 1 mol #O_2# to give 2mol #CO_2#

So:

2 litre #CO# reacts with 1 litre #O_2# to give 2 litre #CO_2#

So 1 litre #CO# reacts with 0.5 litre #O_2# to give 1 litre #CO_2#