Answer:

To solve this problem, we need to use stoichiometry to relate the amount of Fe2S3 to the amount of SO2 produced in the reaction.

First, we need to write and balance the chemical equation:

2Fe2S3 + 9O2 → 2Fe2O3 + 6SO2

The balanced equation shows that 2 moles of Fe2S3 reacts with 6 moles of O2 to produce 6 moles of SO2. Therefore, the molar ratio of Fe2S3 to SO2 is 2:6 or 1:3.

Using this ratio, we can calculate the moles of SO2 produced from the given mass of Fe2S3:

moles of Fe2S3 = 3.8 g / molar mass of Fe2S3

moles of Fe2S3 = 3.8 g / (2 x 55.85 g/mol + 3 x 32.06 g/mol)

moles of Fe2S3 = 0.026 mol

moles of SO2 = moles of Fe2S3 x (6 mol SO2 / 2 mol Fe2S3)

moles of SO2 = 0.026 mol x (6/2)

moles of SO2 = 0.078 mol

Finally, we can calculate the mass of SO2 produced:

mass of SO2 = moles of SO2 x molar mass of SO2

mass of SO2 = 0.078 mol x 64.06 g/mol

mass of SO2 = 5.00 g

Therefore, 5.00 g of SO2 is produced from the reaction of 3.8 g of Fe2S3 and excess O2.

Correct me if I'm wrong.