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For the balanced equation shown below, if 44.9 grams of CoO were reacted with 6.46 grams of O2, how many grams of Co2O3 would be produced?

4CoO+O2=2Co2O3

1.Find the limiting reagent

4CoO + O2

44.9g 6.46g

75g/mol 32g/mol

.598=mol .201=mol

.598 .804

2.Figure out the molar mass of 2Co2O3

Cox2

Ox3

118+48=166

3.Compare the limiting reagent (4CoO) with 2Co2O3 and find ratio

4CoO 2Co2O3

4:2

4.Divide ratio of 4CoO and 2Co2O3

1:.5

5.Take ratio (1:.5) and multiply with the limiting reagent MOL and molar mass of 2Co2O3

1:.5x.598x166

=49.634

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4CoO+O2=2Co2O3

1.Find the limiting reagent

4CoO + O2

44.9g 6.46g

75g/mol 32g/mol

.598=mol .201=mol

__x 1____x 4__.598 .804

**The limiting reagent is 4CoO**2.Figure out the molar mass of 2Co2O3

Cox2

Ox3

118+48=166

3.Compare the limiting reagent (4CoO) with 2Co2O3 and find ratio

4CoO 2Co2O3

4:2

4.Divide ratio of 4CoO and 2Co2O3

1:.5

5.Take ratio (1:.5) and multiply with the limiting reagent MOL and molar mass of 2Co2O3

1:.5x.598x166

=49.634

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