Calculate the chemical amount of aluminum (kmol) required to obtain chromium weighing 5.2

Calculate the chemical amount of aluminum (kmol) required to obtain chromium weighing 5.2 tons from chromium (III) oxide by the aluminothermal method.

1. Let’s write down the equation of the proceeding reaction:

2Al + Cr2O3 = 2Cr + Al2O3;

2.Calculate the chemical amount of chromium (III) oxide:

n (Cr2O3) = m (Cr2O3): M (Cr2O3);

M (Cr2O3) = 2 * Ar (Cr) + 3 * Ar (O) = 2 * 52 + 3 * 16 = 152 g / mol;

n (Cr2O3) = 5200: 152 = 34.21 kmol;

3.Calculate the amount of aluminum entering into the interaction, making a proportion according to the reaction equation:

2 mol of Al reduces 1 mol of Cr2O3;

x kmol – 34.21 kmol Cr2O3;

x = 34.21 * 2: 1 = 68.42 kmol.

Answer: 68.42 kmol.