Question

You need to produce 500.0 g of lithium oxide, using the following unbalanced equation:: Li + O2 Li2O This reaction has an 86% yield. A. Determine the smallest mass of Lithium you would need to start.B. Determine the smallest mass of Oxygen gas you would need to start.

Answer 1

In order to calculate the amount of reactant needed to produce 500g of Li20, we will follow the following steps:

1. Balanced equation:

4Li(s) +O2(g)→2Li2O(s)

2. Determine the theoritical yield of 2Li2O

%yield = {actual yield /theoritical yield }* 100

∴Theoritical yiled ={actual yield / %yield}* 100

={500g 2Li20/86% }*100

∴Theoritical yiled =581 g 2Li2O

3. Determine mass of Lithium and mass of Oxygen:

(3.1)Mass of Lithium,

• given that molecular mass of LiO2 is 38,94 g/mol and Molecular mass Li = 13.88g/mol

mass lithium =581 g Li2O x (1mol Li2O/38.94g)

*4molLi/2molLiO2*13.88g Li/1molLi

=414g of lithium

(3.2) Mass of oxygen

• given that molecular mass of LiO2 is 38,94 g/mol and Molecular mass of oxygen =32g/mol

Mass of Oxygen = 581 g Li2O x (1mol Li2O/38.94g) *

* 1mol O2 /2molLIO2 * 32gO2 /1mol O2

=238 g of Oxygen