How many grams of aluminum oxide containing 8% impurities is required to obtain 500 grams of aluminum nitrate?

Aluminum nitrate Al (NO3) 3 contains one aluminum atom, and aluminum oxide Al2O3 contains two aluminum atoms. To obtain one mole of aluminum nitrate, half a mole of aluminum oxide is required.

Let’s find the amount of a substance contained in 500 grams of aluminum nitrate.

M Al (NO3) 3 = 27 + (14 + 16 x 3) x 3 = 213 grams / mol;

N Al (NO3) 3 = 500/213 = 2.3474 mol;

N Al2O3 = N Al (NO3) 3/2 = 2.3474/2 = 1.1737 mol;

Find the mass of pure aluminum oxide:

M Al2O3 = 27 x 2 + 16 x 3 = 102 grams / mol;

m Al2O3 = 1.1737 x 102 = 119.72 grams;

Find the mass of aluminum oxide containing 8% impurities:

m Al2O3 with impurities = m Al2O3 / 0.92 = 119.72 / 0.92 = 130.13 grams;