Nitrogen and hydrogen combine at high temperature, in the presence of a catalyst, to produce ammonia.
N2(g)+3H2(g)-->2NH3(g)
Assume 4 molecules of nitrogen and 9 molecules of hydrogen are present.
After complete reaction, how many molecules of ammonia are produced?
How many molecules of H2 remain?
How many molecules of N2 remain?
What is the limiting reactant?
.........N2 + 3H2 ==> 2NH3
I........4......9.......0
C
E
If we take 4 N2 that will use up 3*4 = 12 molecules of H2 BUT we don't have 12 molecules; therefore, H2 must be the limiting reagent so in th table we will use H2
...........N2 + 3H2 ==> 2NH3
I..........4.....9.........0
C.........-3....-9........6
E..........1.....0........6
To determine the number of molecules of ammonia produced, we need to find the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
First, we calculate the number of moles for each reactant:
Moles of N2 = 4 molecules / 6.022 x 10^23 molecules/mol = 0.06644 moles
Moles of H2 = 9 molecules / 6.022 x 10^23 molecules/mol = 0.14942 moles
Next, we calculate the number of moles of ammonia that can be produced based on the balanced equation:
Moles of NH3 = 2 moles of ammonia / 1 mole of nitrogen = 2 * 0.06644 moles = 0.13288 moles
Since the stoichiometric coefficient ratio between N2 and NH3 is 1:2, we find that the maximum number of moles of NH3 produced is 0.06644 * 2 = 0.13288 moles.
Now, we need to compare the number of moles of NH3 that can be produced to the actual amount present. Since it is given that 9 molecules of H2 are present, we can calculate the number of moles of NH3 that can be produced from 9 molecules of H2:
Moles of NH3 from H2 = 9 molecules / 6.022 x 10^23 molecules/mol = 0.01493 moles
As this value is less than 0.13288 moles, the number of moles of NH3 that can be produced from the H2 is the limiting reactant.
To find the number of molecules of ammonia produced, we need to convert moles to molecules:
Molecules of NH3 = 0.01493 moles * 6.022 x 10^23 molecules/mol = 8.989 x 10^21 molecules
Therefore, after the complete reaction, 8.989 x 10^21 molecules of ammonia are produced.
To find the remaining number of molecules of H2, we need to subtract the number of moles of H2 used in the reaction from the initial amount:
Remaining moles of H2 = Initial moles of H2 - Moles of NH3 from H2 = 0.14942 moles - 0.01493 moles = 0.13449 moles
Molecules of H2 = 0.13449 moles * 6.022 x 10^23 molecules/mol = 8.097 x 10^22 molecules
Therefore, after the complete reaction, 8.097 x 10^22 molecules of H2 remain.
To find the remaining number of molecules of N2, we need to consider the stoichiometric coefficient ratio between N2 and NH3:
Moles of N2 reacted = Moles of NH3 from H2 * 1/2 (stoichiometric ratio) = 0.01493 moles * 1/2 = 0.007465 moles
Remaining moles of N2 = Initial moles of N2 - Moles of N2 reacted = 0.06644 moles - 0.007465 moles = 0.058975 moles
Molecules of N2 = 0.058975 moles * 6.022 x 10^23 molecules/mol = 3.550 x 10^22 molecules
Therefore, after the complete reaction, 3.550 x 10^22 molecules of N2 remain.
The limiting reactant is the one that is completely consumed in the reaction, which in this case is H2.
To find the number of molecules of ammonia produced, we first need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be produced.
Step 1: Convert the given number of molecules of nitrogen and hydrogen into moles.
- 4 molecules of nitrogen = 4/6.022 x 10^23 moles of nitrogen (Avogadro's number is 6.022 x 10^23)
- 9 molecules of hydrogen = 9/6.022 x 10^23 moles of hydrogen
Step 2: Determine the stoichiometric ratio between the reactants and the products.
From the balanced chemical equation, we can see that the mole ratio between nitrogen and ammonia is 1:2, and the mole ratio between hydrogen and ammonia is 3:2.
Step 3: Determine the limiting reactant.
To find the limiting reactant, we need to compare the number of moles of nitrogen and hydrogen with their stoichiometric ratios.
- Nitrogen: 4/6.022 x 10^23 moles = 6.64 x 10^-24 moles
- Hydrogen: 9/6.022 x 10^23 moles = 1.49 x 10^-23 moles
To compare the two, we need to multiply the moles by the stoichiometric ratios:
- Nitrogen: 6.64 x 10^-24 moles x 2/1 = 1.33 x 10^-23 moles of ammonia possible
- Hydrogen: 1.49 x 10^-23 moles x 2/3 = 9.92 x 10^-24 moles of ammonia possible
Since the number of moles of ammonia possible from nitrogen is lower, nitrogen is the limiting reactant.
Step 4: Calculate the number of molecules of ammonia produced.
Since 1 mole of ammonia corresponds to 6.022 x 10^23 molecules of ammonia (Avogadro's number), we can find the number of ammonia molecules produced.
- Molecules of ammonia produced = 1.33 x 10^-23 moles x 6.022 x 10^23 molecules/mole = 8.01 x 10^4 molecules
Step 5: Calculate the amount of excess hydrogen molecules.
From the stoichiometric ratio, we know that 3 moles of hydrogen produce only 2 moles of ammonia. Therefore, the excess hydrogen molecules can be calculated as follows:
- Excess hydrogen = (1.49 x 10^-23 moles - 1.33 x 10^-23 moles) x 6.022 x 10^23 molecules/mole = 9.60 x 10^3 molecules
Step 6: Calculate the amount of excess nitrogen molecules.
Since nitrogen is the limiting reactant, all of it has reacted completely. Therefore, no nitrogen molecules remain.
In summary:
- The number of molecules of ammonia produced = 8.01 x 10^4 molecules
- The number of molecules of H2 remain = 9.60 x 10^3 molecules
- The number of molecules of N2 remain = 0 molecules
- The limiting reactant is nitrogen (N2).
Molecules of H2: 3 molecules.
Nitrogen atoms present: 3 atoms
Moles of NH3 formed: 2 moles