# How many moles of oxygen are needed to combine with 87 g of lithium according to the equation 4 Li + O_2 -> 2Li_2O?

Dec 21, 2015

Moles of Li? Approx. 12 mol. Therefore approx. 3 moles of oxygen gas are required.

#### Explanation:

$4 L i + {O}_{2} \rightarrow 2 L {i}_{2} O$

You have the stoichiometric equation, and we ASSUME excess oxygen.

So moles of metal $=$ $\frac{87 \cdot \cancel{g}}{6.94 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $=$ $12.5$ $m o l$.

Therefore, we require a stoichiometric quantity of oxygen gas, and you have kindly provided the equation:

$\frac{\left(12.58 \cdot \cancel{m o l}\right) \cdot \cancel{L i}}{\left(4 \cancel{m o l} \cdot {O}_{2}\right) \left(m o {l}^{- 1}\right) \cancel{L i}}$ $=$ $3.2$ $m o l$ ${O}_{2}$ gas.

Note that we deal with oxygen gas, the diatomic molecule.