Calculate the mass of zinc oxide that can be obtained from 485 kg of zinc blende ZnS with w (ZnS) = 80%

Let’s implement the solution:
1. In accordance with the condition of the problem, we compose the equation:
2ZnS + 3O2 = 2ZnO + 2SO2 – OBP, zinc oxide, sulfur dioxide are released;
2. We make calculations:
M (ZnS) = 97.3 g / mol;
M (ZnO) = 79.5 g / mol.
3. Determine the theoretical mass of zinc blende (zinc sulfide):
W = m (practical) / m (theoretical) * 100;
m (practical) = 485 kg. = 485,000 g;
m (theoretical) ZnS = 485000 / 0.80 = 606250 g = 606 kg. 250 g
4. Calculate the number of moles of the starting material:
Y (ZnS) = m / M = 606 250 / 97.3 = 6230.7 mol.
5. Proportion:
6230.7 mol (ZnS) – x mol (ZnO);
-2 mol                    – 2 mol, that is, the amount of these substances according to the equation is equal to 2 mol.
Y (ZnO) = 6230.7 mol.
6. Find the mass of the product:
m (ZnO) = Y * M = 6230.7 * 79.5 = 495 kg. 34 g
Answer: during the roasting of zinc sulfide, zinc oxide weighing 495 kg was obtained. 34 g