Balance the equation and determine how many liter of nitrogen are required to react with 7.3 L of hydrogen to produce ammonia (n
1 answer:
Answer:
4.867 L of ammonia
Explanation:
Using Haber's process to form ammonia using Nitrogen and hydrogen, the equation is :
N₂ + 3H₂ → 2NH₃
Here, 3 moles of hydrogen gas gives 2 moles of ammonia.
1 mole of any substance occupies 22.4L at STP
So, 3 x 22.4L of hydrogen gives 2 x 22.4 L of ammonia
Then 7.3 L of hydrogen will give:
=
=
= 4.867 L of ammonia
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Answer: The percent yield of this combination reaction is 41.3 %
Explanation : Given,
Mass of = 5.0 g
Mass of = 10.0 g
Molar mass of = 23 g/mol
Molar mass of = 71 g/mol
First we have to calculate the moles of and
.
and,
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation will be:
From the balanced reaction we conclude that
As, 2 mole of react with 1 mole of
So, 0.217 moles of react with
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of
From the reaction, we conclude that
As, 2 mole of react to give 2 mole of
So, 0.217 mole of react to give 0.217 mole of
Now we have to calculate the mass of
Molar mass of = 58.5 g/mole
Now we have to calculate the percent yield of this reaction.
Percent yield =
Actual yield = 5.24 g
Theoretical yield = 12.7 g
Percent yield =
Percent yield = 41.3 %
Therefore, the percent yield of this combination reaction is 41.3 %