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Scilla [17]
3 years ago
11

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 pr

oduced?
FeS2 + O2 = Fe2O3 + SO2
Chemistry
2 answers:
podryga [215]3 years ago
6 0
 <span>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.</span>
bija089 [108]3 years ago
3 0

Answer:

Amount of Fe2O3 produced is 197 g

Explanation:

Step 1: Deduce the limiting reactant

Mass of FeS2 = 294 g

Molar Mass of FeS2 = 120 g/mol

Moles\ FeS2 = \frac{Mass}{Molar Mass} =\frac{294}{120} =2.45 moles

Mass of O2 = 176 g

Molar mass of O2 = 32 g/mol

Moles\ O2 = \frac{Mass}{Molar Mass} =\frac{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

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