Question

0.02 mol of aluminum is is burned in oxygen and the product is reacted with 2.00 mol dm^-3 hydrochloric acid.
what is the minimum volume of acid that will be required for complete reaction?

Answer 1

There are two reactions in this task:

4Al + 3O₂ = 2Al₂O₃

Al₂O₃ + 6HCl = 2AlCl₃ + 3H₂O

0.02 mol of aluminum can produce X mol of Al₂O₃

0.02 X
4Al + 3O₂ = 2Al₂O₃
4
2

X = 0.02*2/4 = 0.01 mol
Than 0.01 mol of Al₂O₃ can react with Y mol of HCl

0.01 Y
Al₂O₃ + 6HCl = 2AlCl₃ + 3H₂O
1 6

Y = 0.06 mol

Concentration of given solution is 2 M
It means that there is 2 mol in one L, and you need to know the volume if
you have 0.06 mol

2 mol ---------- 1 L
0.06 mol------- V

V = 0.06*1/2 = 0.03 L (30 ml)