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For the balanced equation shown below, if 14.5 grams of C2H3F3 were reacted with 3.00 grams of O2, how many grams of HF would be produced?
C2H3F3+O2=2CO+3HF
1.Find the limiting reagent
C2H3F3 + O2
14.5g 3.00g
84g/mol 32g/mol
.172=mol .093=mol
x 1 x 1
.172 .093
The limiting reagent is O2
2.Figure out the molar mass of 3HF
Hx1
Fx1
1+19=20
3.Compare the limiting reagent (O2) with 3HF and find ratio
O2 3HF
1:3
4.Divide ratio of O2 and 3HF
1:3
5.Take ratio (1:3) and multiply with the limiting reagent MOL and molar mass of 3HF
1:3x.093x20
=5.58
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C2H3F3+O2=2CO+3HF
1.Find the limiting reagent
C2H3F3 + O2
14.5g 3.00g
84g/mol 32g/mol
.172=mol .093=mol
x 1 x 1
.172 .093
The limiting reagent is O2
2.Figure out the molar mass of 3HF
Hx1
Fx1
1+19=20
3.Compare the limiting reagent (O2) with 3HF and find ratio
O2 3HF
1:3
4.Divide ratio of O2 and 3HF
1:3
5.Take ratio (1:3) and multiply with the limiting reagent MOL and molar mass of 3HF
1:3x.093x20
=5.58
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