Consider the combustion of methane: Ch4(g)+2O2(g) --> CO2(g)+2H2O(g), suppose 2.8 moles of methane are allowed to react with 3 moles of oxygen, what is the limiting reactant?

1 Answer
Oct 28, 2015

Answer:

#O_2# is the limiting reactant.

Explanation:

There are different methods of finding the limiting reactant, here is the simplest to my opinion.

The reaction is:
#CH_4(g) + color(blue)(2)O_2(g)->CO_2(g)+2H_2O(g)#

Find the molar ratio between the experimental number of moles and theoretical number of moles of the reactants:

For #CH_4#: #(2.8 cancel(mol))/(1 cancel(mol))=2.8#.
The #1 " mol"# is taking from the coefficient of #CH_4# from the balanced equation.

For #O_2#: #(3 cancel(mol))/(color(blue)(2 cancel(mol)))=1.5#.
The #2 " mol"# is taking from the coefficient of #O_2# from the balanced equation.

The reactant that gives the smaller molar ratio is the limiting reactant which is in this case the oxygen #O_2#

More Explanation:
To consume the #2.8 " moles of " CH_4# we need #5.6 " moles of " O_2# since the molar ratio is 1:2. We have only #3 " moles of " O_2#; therefore, #O_2# is the limiting reactant.