if your actual yield in an experiment was 20.0 grams of H2O, what would be your % yield?
Answers
Answer:
Once again, start with the balanced chemical equation for the combustion of methane,
C
H
4
C
H
4
+
2
O
2
→
C
O
2
+
2
H
2
O
You have a
1:2
mole ratio between methane and water - this means that the number of moles of water produced must be twice the number of moles of methen that reacted.
The theoretical yield is the yield you'd get for a 100% reaction - i.e. when all the methane reacted and water was produced according to the aforementioned mole ratio. So, in theory, this reaction would produce
1.60 g methane
⋅
1 mole methane
16.0 g
⋅
2 moles water
1 mole methane
⋅
18.0 g
1 mole water
=
3.60 g water
Now, the experiment you ran produced
3.30 g
of water; this means that not all the methane reacted
→
oxygen was the limiting reagent.
The reaction's percent yield for water will be
%yield
=
experimental yield
actual yield
⋅
100
%yield water
=
3.30 g
3.60 g
⋅
100
=
91.7%
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