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

1 Answer
anor277
Dec 21, 2015

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.

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