How many moles of oxygen are needed to combine with 87 g of lithium according to the equation $4 L i + O_{2} \to 2 L i_{2} O$ $4 Li + O_2 -> 2Li_2O$ ?

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Answer 1 · 2015-12-21T09:10:43

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

$4 L i + O_{2} \to 2 L i_{2} O$ $4Li + O_2 rarr 2Li_2O$

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.

How many moles of oxygen are needed to combine with 87 g of lithium according to the equation 4 Li + O_2 -> 2Li_2O4Li+O2→2Li2O4 Li + O_2 -> 2Li_2O?