Question

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?

Answer 1

`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.