Question

A mixture of Al2O3 and CuO weighing 18.371 mg was heated under H2 at 1000 C to give 17.462 mg of Al2O3 +Cu. The other product is H2O. Find wt% Al2O3 in the original mixture.

thanks

Answer 1

m total = 18.371

Al2O3 + CuO goes to Al2O3 + Cu

Oxygen is lost

difference of oxygen

m Oxygen = 18.371-17.462 = 0.909mg of Oxygen are gone

mol = masS/MW = (0.909*10^-3)/16= 0.0000568125 mol of Oxygen

then, according to stoichiometry

1 mol of O contains 1 mol of Cu

then

0.0000568125 mol of O contains 0.0000568125 mol of CuO

MW CuO = 79.545

mass = mol*MW = 0.0000568125*79.545 = 0.00451915031 = 4.51915 mg of CuO

then

Mg of AL2O3 = 18.371-4.51915 = 13.852 mg of Al2O3

then

% Al2O3 = mass of Al2O3 / total mass * 100 = 13.852/(18.371 ) * 100 = 75.4014 %