What amount of iron must be taken to obtain 200 g of iron sulfide, if its practical yield is 80%

What amount of iron must be taken to obtain 200 g of iron sulfide, if its practical yield is 80% in comparison with the theoretically possible.

Let’s solve the problem, make up the equation:
Fe + S = FeS (2) – compound reaction, iron sulfide was obtained;
Let’s make calculations using the formulas of substances:
M (Fe) = 55.8 g / mol;
M (FeS) = 87.8 g / mol;
Determine the theoretical mass of iron sulfide, if the value of W = 80% is known
W = m (practical) / m (theoretical) * 100;
m (theoretical) = 0.80 * 87.8 = 70.24 g.
We find the number of moles of iron sulfide salt according to the formula:
Y (FeS) = m / M = 70.24 / 87.8 = 0.8 mol;
According to the equation, the amount of moles of iron and the reaction product of iron sulfide are equal to 1 mol, which means that Y (Fe) = 0.8 mol.
Answer: The number of moles of iron is 0.8 moles.