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For the balanced equation shown below, if 57.5 grams of Fe2S2 were reacted with 23.8 grams of O2, how many grams of Fe2O3 would be produced?

4Fe2S2+11O2=2Fe2O3+8SO2

1.Find the limiting reagent

4Fe2S2 + 11O2

123g/mol 32g/mol

.467=mol .743=mol

5.137 2.972

2.Figure out the molar mass of 2Fe2O3

Fex2

Ox3

118+48=166

3.Compare the limiting reagent (11O2) with 2Fe2O3 and find ratio

11O2 2Fe2O3

11:2

4.Divide ratio 11O2 and 2Fe2O3

1:.181

5.Take ratio (1:.909) and multiply with the limiting reagent MOL and molar mass of 2Fe2O3

1:.181x.743x166

=22.324

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4Fe2S2+11O2=2Fe2O3+8SO2

1.Find the limiting reagent

4Fe2S2 + 11O2

__57.5g____23.8g__123g/mol 32g/mol

.467=mol .743=mol

__x 11____x 4__5.137 2.972

**The limiting reagent is 11O2**2.Figure out the molar mass of 2Fe2O3

Fex2

Ox3

118+48=166

3.Compare the limiting reagent (11O2) with 2Fe2O3 and find ratio

11O2 2Fe2O3

11:2

4.Divide ratio 11O2 and 2Fe2O3

1:.181

5.Take ratio (1:.909) and multiply with the limiting reagent MOL and molar mass of 2Fe2O3

1:.181x.743x166

=22.324

Still don't understand?

Click Here to look at step by step examples

Click Here to watch a video on how to find the limiting reactant