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For the balance equation shown below, if 81.1 grams of C2H4 were reacted with 180 grams of O2 how many grans of H2O would be produced?
C2H4+3O2=2CO2+2H2O
1.Find limiting reagent
C2H4 + 3O2
81.1g 180g
28g/mol 32g/mol
2.89=mol 5.62=mol
x 3 x 1
8.67 5.62
The limiting reagent is 3O2
2.Figure out the molar mass of 2H2O
Hx2
OX1
2+16=18
3.Compare the limiting reagent (3O2) with 2H2O and find ratio
3O2 2H2O
3:2
4.Divide ratio of 3O2 and 2H2O
1:.66
5.Take ratio (1:.66) and multiply with the limiting reagent MOL and molar mass of 2H2O
1:.66x5.62x18
=66.7656
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C2H4+3O2=2CO2+2H2O
1.Find limiting reagent
C2H4 + 3O2
81.1g 180g
28g/mol 32g/mol
2.89=mol 5.62=mol
x 3 x 1
8.67 5.62
The limiting reagent is 3O2
2.Figure out the molar mass of 2H2O
Hx2
OX1
2+16=18
3.Compare the limiting reagent (3O2) with 2H2O and find ratio
3O2 2H2O
3:2
4.Divide ratio of 3O2 and 2H2O
1:.66
5.Take ratio (1:.66) and multiply with the limiting reagent MOL and molar mass of 2H2O
1:.66x5.62x18
=66.7656
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 reactants