What is the mass of the metal obtained by the motor-drive for its reduction from Fe2O3 oxide weighing 20 g;

What is the mass of the metal obtained by the motor-drive for its reduction from Fe2O3 oxide weighing 20 g; 10% impurities if the mass fraction of the reaction product yield is 90% of the theoretically possible?

Given:
m ore (Fe2O3) = 20 g
ω approx. = 10%
ω out. = 90%

To find:
m pract. (Fe) -?

Decision:
1) Write the reaction equation:
Fe2O3 + 3H2 => 2Fe + 3H2O;
Fe2O3 + 3СО => 2Fe + 3СО2;
Fe2O3 + 3C => 2Fe + 3CO;
By the condition, it is not said which reaction is the reduction of Fe. However, this is not so important: it is important that in all reactions the ratio Fe2O3: Fe = 1: 2;

2) Find the mass fraction of Fe2O3 in the sample:
ω (Fe2O3) = 100% – ω approx. = 100% – 10% = 90%;

3) Find the mass of pure Fe2O3:
m (Fe2O3) = ω (Fe2O3) * m ore (Fe2O3) / 100% = 90% * 20/100% = 18 g;

4) Find the amount of the substance Fe2O3:
n (Fe2O3) = m (Fe2O3) / Mr (Fe2O3) = 18/160 = 0.1 mol;

5) Find the amount of the substance Fe (taking into account the reaction equation):
n (Fe) = n (Fe2O3) * 2 = 0.1 * 2 = 0.2 mol;

6) Find the theoretical mass of Fe:
m theor. (Fe) = n (Fe) * Mr (Fe) = 0.2 * 56 = 11.2 g;

7) Find the practical mass of Fe:
m pract. (Fe) = ω out. * m theor. (Fe) / 100% = 90% * 11.2 / 100% = 10 g.

Answer: The practical weight of Fe is 10 g.