## :( INCORRECT :(

For the balanced equation shown below, if 62.2 grams of C4H8O2 were reacted with 212 grams of O2, how many grams of H2O would be produced?

C4H8O2+5O2=4CO2+4H2O

1.Find the limiting reagent

C4H8O2 + 5O2

88g/mol 32g/mol

.706=mol 6.625=mol

3.53 6.625

2.Figure out the molar mass of 4H2O

Hx2

Ox1

2+16=18

3.Compare the limiting reagent (C4H8O2) with 4H2O and find ratio

C4H8O2 4H2O

1:4

4.Divide ratio of C4H8O2 and 4H2O

1:4

5.Take ratio (1:4) and multiply with the limiting reagent MOL and molar mass of 4H2O

1:4x.706x18

=50.832

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

C4H8O2+5O2=4CO2+4H2O

1.Find the limiting reagent

C4H8O2 + 5O2

__62.2g____212g__88g/mol 32g/mol

.706=mol 6.625=mol

__x 5____x 1__3.53 6.625

**The limiting reagent is C4H8O2**2.Figure out the molar mass of 4H2O

Hx2

Ox1

2+16=18

3.Compare the limiting reagent (C4H8O2) with 4H2O and find ratio

C4H8O2 4H2O

1:4

4.Divide ratio of C4H8O2 and 4H2O

1:4

5.Take ratio (1:4) and multiply with the limiting reagent MOL and molar mass of 4H2O

1:4x.706x18

=50.832

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