What mass of Cr2O3 containing 10% impurities is necessary to obtain 52 gr. Cr, aluminothermic

What mass of Cr2O3 containing 10% impurities is necessary to obtain 52 gr. Cr, aluminothermic method Cr2O3 + Al = Al2O3 + Cr.

Let’s execute the solution:

According to the condition of the problem, we write:
Xr -?; 10% – impurities m = 52 g

Cr2O3 + 2Al = Al2O3 + 2Cr – OBP, chromium obtained;

2. Calculations by formulas:

M (Cr2O3) = 151.8 g / mol.

M (Cr) = 51.9 g / mol.

Y (Cr) = m / M = 52/51, 9 = 1 mol.

3. Proportion:

X mol (Cr2O3) – 1 mol (Cr);

-1 mol -2 mol hence, X mol (Cr2O3) = 1 * 1/2 = 0.5 mol.

4. Find the mass of the original substance:

m (Cr2O3) = Y * M = 0.5 * 151.8 = 75.9 g;

m (Cr2O3) = 75.9 * (1 – 0.10) = 68.3 g (mass without impurities).

Answer: the mass of chromium oxide is 68.3 g