Question

If 294 grams of FeS2 is allowed ti react with 176 grams of O2 according to the following equation how many grams of Fe2O3 are produced?

FeS2 + O2 = Fe2O3 + SO2

Answer 1

Amount of Fe2O3 produced is 197 g

Step-by-step explanation:

Step 1: Deduce the limiting reactant

Mass of FeS2 = 294 g

Molar Mass of FeS2 = 120 g/mol

`Moles\ FeS2 = (Mass)/(Molar Mass) =(294)/(120) =2.45 moles`

Mass of O2 = 176 g

Molar mass of O2 = 32 g/mol

`Moles\ O2 = (Mass)/(Molar Mass) =(176)/(32) =5.5 moles`

Since moles of FeS2 < O2, then FeS2 is the limiting reactant which will dictate the amount of product formed

Step 2: Calculate the mass of Fe2O3 produced

The balanced equation is:

4FeS2 + 11O2 → 2Fe2O3 + 8SO2

Based on the reaction stoichiometry:

4 moles of FeS2 produced 2 moles of Fe2O3

Therefore, 2.45 moles of FeS2 will produce 1.23 moles Fe2O3

Molar mass of Fe2O3 = 160 g/mol

`Mass\ of\ Fe2O3 = moles * molar\ mass = 1.23 * 160 = 196.8 g`

Answer 2

You must balance your equation correctly.
Here is your answer:

294gFeS2 x 1molFeS2/119.99 x 11mols O2/4mols FeS2--> 6.738mol O2

176gO2 x 1mol O2/32gO2 x 4mols FeS2/11mol FeS2--> 2mols FeS2
Now choose the molecule with the lowest amount (Limiting Reagent)

2molsFeS2 x 2molsFe2O3/4molsFeS2 x 159.7g
159.7g Fe2O3 grams produced.