4FeS2 + 11O2 > 2Fe2O3 + 8SO2

Given 75.0 grams of iron (IV) sulfide, how many grams of sulfur dioxide are produced?

80.13g of SO2

Explanation:

4FeS2 + 11O2 —> 2Fe2O3 + 8SO2

Molar Mass of FeS2 = 55.8 + (2x32) = 119.8g/mol

Mass of FeS2 from the balanced equation = 4 x 119.8 = 479.2

Molar Mass of SO2 = 32 + (16x2) = 32 + 32 = 64g/mol

Mass of SO2 from the balanced equation = 8 x 64 = 512g

From the equation,

479.2g of FeS2 produced 512g of SO2.

Therefore, 75g of FeS2 will produce = (75 x 512)/479.2 = 80.13g of SO2

h let me think

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