Respuesta :
Answer
19.7971 g Li₂O
Procedure
To solve this question we first need to verify if the equation is balanced. In this case, it is already balanced.
4Li(s)+O2(g)→2Li₂O(s)
Then we will need to convert the grams of both reagents to moles using the molecular weight, in order to use the stoichiometry of the reaction.
[tex]15.7\text{ g Li}\frac{1\text{ mol Li}}{6.941\text{ g Li}}=2.2619\text{ mol Li}[/tex][tex]10.6\text{ g O}_2\frac{1\text{ mol O}_2}{31.99\text{ g O}_2}=0.3313\text{ mol O}_2[/tex]Then we determine the limiting reagent by stoichiometry as follows:
[tex]2.2619\text{ mol Li}\frac{2\text{ mol Li}_2\text{O}}{4\text{ mol Li}}=1.13095\text{ mol Li}_2\text{O}[/tex][tex]0.3313\text{ mol O}_2\frac{2\text{ mol Li}_2\text{O}}{1\text{ mol O}_2}=\text{ 0.6626 mol Li}_2\text{O}[/tex]Given that the lowest value comes from molecular oxygen, this will be the limiting reagent.
Then we use the value from the molecular oxygen and convert it into grams.
[tex]0.6626\text{ mol Li}_2\text{O}\frac{1\text{ mol Li}_2\text{O}}{29.8814\text{ Li}_2\text{O}}=\text{ 19.7971 g Li}_2\text{O}[/tex]