Answer to Question #177023 in General Chemistry for Elijah

Question

You start with 5.4 g of Al and 5.12 g of O². What is the limited reactant

Answer 1

Al + O2 -> Al2O3

4Al + 3O2 --> 2Al2O3

moles Al present = 5.4 g Al x 1 mole Al/26.98 g

= 0.2001 mol of Al

Mole of O2 present = 5.12/31.998

= 0.16 mol of O2

Dividing each reactant by it's coefficient in the balanced equation one obtains:

Al = 0.2001/4 = 0.050025

O2= 0.16/ 3 = 0.05333

Hence Al is the limiting reactant