Question

What is the limiting reactant when 8.4 moles of lithium react with 4.6 moles of oxygen gas?

Answer 1

To determine the limiting reactant, we divide the number of moles to its corresponding stoichiometric coefficient. Balancing the equation above, we get 4 Li + O2 = 2Li2O. For lithium, the result is 2.1 and for oxygen, the result is 4.6. The limiting reactant has the smaller ratio. So, the limitin reactant is lithium.

Answer 2

Answer: Lithium

Explanation: The balanced chemical equation is:

$4Li+O_2\rightarrow 2Li_2O$

It can be seen, 4 moles of lithium combines with 1 mole of oxygen gas to produce 2 moles of lithium oxide.

Thus 8.4 moles of lithium combines with=$\frac{1}{4}\times {8.4}=2.1moles$of oxygen gas to produce 4.2 moles of lithium oxide.

As, Lithium limits the formation of product, it is the limiting reagent and Oxygen gas is present in excess, it is called the excess reagent. (4.6-2.1)=2.5 moles of oxygen gas are present in excess.