4 FeS2+ 11 O2 → 2 Fe2O3+ 8SO2

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

75g of FeS2 will produce 80.1g of SO2.

Explanation:

4FeS2 + 11O2 > 2Fe2O3 + 8SO2

Step 1: obtain the number of mole to produce the gas SO2

4 moles of FeS2 produces 8 moles of SO2

1 mole of FeS2 will produce 2 moles of SO2

Step 2:

Find the relative molecular mass (RMM) of FeS2 and SO2

(Fe= 55.8, S = 32)

RMM of FeS2= 119.8g/mol

RMM of SO2 = 64g/mol

Step 3:

Equate the RMM into the mole equation

1mole of FeS2 = 2 moles of SO2

119.8g/mol = 64*2= 128g/mol of SO2

75g will produce ( 128 * 75 / 119.8)

= 80.1g

So therefore, 75g of FeS2 will produce 80.1g of SO2.

D) 80.0 g

I did the assignment already.

the answer is a! (2.9 x 10^-18 j) hope i could someone. : )