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?
1
Expert's answer
2013-01-28T09:40:12-0500
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
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