Answer to Question #163060 in General Chemistry for Dawson

Question

In a combination reaction, 1.62 g of lithium is mixed with 6.87 g of oxygen.

Answer a, b, and c. I am lost.

a) Which reactant is present in excess? 

 b) How many moles of the product are formed?  

c) After the reaction, How many grams of each reactant and product are present? 

Blank g Li 

Blank g O2 

Blank g LI2O

Answer 1

4Li+O₂=2Li₂O

n(Li)=m/M=1.62/6.94=0.233mol

n(O₂)=m/M=6.87/32=0.214mol

0.233/4=0.0582mol

a)in excess O₂

b) n(Li₂O)=0.233/2=0.1165moles

c)m(Li)=0g

m(O2)=0g

m(Li2O)=n*M=0.1165*29.88=3.48g