Question

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

Answer 1

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

Explanation:

`4Li + O_2 rarr 2Li_2O`

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

So moles of metal `=` `(87*cancelg)/(6.94*cancelg*mol^-1)` `=` `12.5` `mol`.

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

`((12.58*cancel(mol))*cancel(Li))/((4cancel(mol)*O_2)(mol^(-1))cancel(Li))` `=` `3.2` `mol` `O_2` gas.

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