Question

For each of the following reactions, calculate the grams of indicated product when 15.7 g of the first reactant and 10.6 g of the second reactant are used.4Li(s)+O2(g)→2Li2O(s) Calculate the mass of Li2O.

Answer 1

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.

`15.7\text{ g Li}\frac{1\text{ mol Li}}{6.941\text{ g Li}}=2.2619\text{ mol Li}`
`10.6\text{ g O}_2\frac{1\text{ mol O}_2}{31.99\text{ g O}_2}=0.3313\text{ mol O}_2`

Then we determine the limiting reagent by stoichiometry as follows:

`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}`
`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}`

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.

`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}`