Answer to Question #22874 in Inorganic Chemistry for fari

Question #22874
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?
Expert's answer
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

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)

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Assignment Expert
25.02.14, 18:16

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21.02.14, 17:01

Thank you very much !

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