Nitrogen and hydrogen react to form ammonia. N2+3H2=2NH3. 12 dm3 of hydrogen reacted with excess nitrogen to form 2 dm3 of ammonia. What is the percentage yield of ammonia at room temperature and pressure?
Answers
Answer:
75%
Explanation:
The first thing that you need to do here is to calculate the theoretical yield of the reaction, i.e. what you get if the reaction has a
100
%
yield.
The balanced chemical equation
N
2
(
g
)
+
3
H
2
(
g
)
→
2
NH
3
(
g
)
tells you that every
1
mole of nitrogen gas that takes part in the reaction will consume
3
moles of hydrogen gas and produce
1
mole of ammonia.
In your case, you know that
1
mole of nitrogen gas reacts with
1
mole of hydrogen gas. Since you don't have enough hydrogen gas to ensure that all the moles of nitrogen gas can react
what you need
3 moles H
2
>
what you have
1 mole H
2
you can say that hydrogen gas will act as a limiting reagent, i.e. it will be completely consumed before all the moles of nitrogen gas will get the chance to take part in the reaction.
So, the reaction will consume
1
mole of hydrogen gas and produce
1
mole H
2
⋅
2 moles NH
3
3
moles H
2
=
0.667 moles NH
3
at
100
%
yield. This represents the reaction's theoretical yield.
Now, you know that the reaction produced
0.50
moles of ammonia. This represents the reaction's actual yield.
In order to find the percent yield, you need to figure out how many moles of ammonia are actually produced for every
100
moles of ammonia that could theoretically be produced.
You know that
0.667
moles will produce
0.50
moles, so you can say that
100
moles NH
3
.
in theory
⋅
0.50 moles NH
3
.
actual
0.667
moles NH
3
.
in theory
=
75 moles NH
3
.
actual
Therefore, you can say that the reaction has a percent yield equal to
% yield = 75%
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Answer:
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