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
Dr. Hayek
Oct 28, 2015

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.

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