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For the balanced equation shown below, if 10.4 grams of C3H6 were reacted with 36.0 grams of O2, how many grams of CO would be produced?
C3H6+3O2=3CO+3H2O
1.Find the limiting reagent
C3H6 + 3O2
10.4g 36.0g
42g/mol 32g/mol
.247=mol 1.125=mol
x 3 x 1
.741 1.125
The limiting reagent is C3H6
2.Figure out the molar mass of 3CO
Cx1
Ox1
12+16=28
3.Compare the limiting reagent (C3H6) with 3CO and find the ratio
C3H6 3CO
1:3
4.Divide ratio of C3H6 and 3CO
1:3
5.Take ratio (1:3) and multiply with the limiting reagent MOL and molar mass of 3CO
1:3x.247x28
=20.748
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C3H6+3O2=3CO+3H2O
1.Find the limiting reagent
C3H6 + 3O2
10.4g 36.0g
42g/mol 32g/mol
.247=mol 1.125=mol
x 3 x 1
.741 1.125
The limiting reagent is C3H6
2.Figure out the molar mass of 3CO
Cx1
Ox1
12+16=28
3.Compare the limiting reagent (C3H6) with 3CO and find the ratio
C3H6 3CO
1:3
4.Divide ratio of C3H6 and 3CO
1:3
5.Take ratio (1:3) and multiply with the limiting reagent MOL and molar mass of 3CO
1:3x.247x28
=20.748
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