What volume of sulfur dioxide will be released when burning 12 g of pyrite if the product yield is 90%?

In accordance with the condition of the problem, we will compose an equation, we will carry out the arrangement of the coefficients:
4FeS2 + 11O2 = 8SO2 + 2Fe2O3 – redox reaction, sulfur oxide was obtained (4);
Let’s calculate the molar masses of substances:
M (FeS2) = 119.8 g / mol; M (SO2) = 64 g / mol;
Determine the amount of moles of pyrite:
Y (FeS2) = m / M = 12 / 119.8 = 0.1 mol.
Let’s make the proportion:
0.1 mol (FeS2) – X mol (SO2);
-4 mol -8 mol hence, X mol (SO2) = 0.1 * 8/4 = 0.2 mol;
Let’s calculate the volume of sulfur dioxide:
V (SO2) = 0.2 * 22.4 = 4.48 liters.
We determine the practical volume if the product yield is known to be 90%.
V (practical) = 0.90 * 4.48 = 4.03 liters.
Answer: The volume of sulfur oxide is 4.03 liters.