Question

Suppose 2.8 moles of methane are allowed to react with 5 moles of oxygen.

Answer 1

Answer : The limiting reagent is $O_2$

Solution : Given,

Moles of methane = 2.8 moles

Moles of $O_2$ = 5 moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that

As, 2 mole of $O_2$ react with 1 mole of $CH_4$

So, 5 moles of $O_2$ react with $\frac{5}{2}=2.5$ moles of $CH_4$

From this we conclude that, $CH_4$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Hence, the limiting reagent is $O_2$