A mixture of nitrogen gas and hydrogen gas reacts in a closed container to form ammonia
gas. The reaction ceases before either reactant is completely consumed. At this stage the
container has 3.0 mol each of all three gases; nitrogen, hydrogen and ammonia. How many
moles of hydrogen and nitrogen were present initially?
Ah, chemistry! The realm of atoms and reactions. Let's see if I can clown around with these numbers for you.
Since all three gases have 3.0 mol in the container at the end, it means that at the beginning of the reaction, the amount of nitrogen and hydrogen combined was also 3.0 mol. That's all the gas we got!
But since we know that the nitrogen and hydrogen reacted to form ammonia, we can assume that they combined in a 1:3 ratio (1 mol of nitrogen reacts with 3 mol of hydrogen). Keeping the ratio in mind, it means there were 1.0 mol of nitrogen and 2.0 mol of hydrogen initially. Fun times, right?
So, to sum it up, there were 1.0 mol of nitrogen and 2.0 mol of hydrogen present initially. Now I hope these numbers entertained you as much as I enjoyed juggling them!
To find out how many moles of hydrogen and nitrogen were present initially, we need to set up the balanced chemical equation for the reaction between nitrogen gas and hydrogen gas to form ammonia gas:
N2 + 3H2 → 2NH3
From the equation, we can see that 1 mole of nitrogen reacts with 3 moles of hydrogen to form 2 moles of ammonia.
Given that at the end of the reaction there are 3.0 moles of each gas, we can set up the following equation to represent the moles of each component:
N2 + 3H2 = 2NH3
Let's say the initial moles of nitrogen and hydrogen are x and y, respectively.
Initially, we have x moles of nitrogen and y moles of hydrogen, and after the reaction, we have x - 2 moles of nitrogen (remaining), y - 6 moles of hydrogen (remaining), and 3 moles of ammonia.
Based on the given information, we have:
x - 2 = 3 (moles of nitrogen remaining)
y - 6 = 3 (moles of hydrogen remaining)
Solving these equations will give us the initial moles of nitrogen (x) and hydrogen (y).
x - 2 = 3
x = 3 + 2
x = 5
y - 6 = 3
y = 3 + 6
y = 9
Therefore, the initial amount of nitrogen was 5 moles, and the initial amount of hydrogen was 9 moles.
To find the initial number of moles of nitrogen and hydrogen, we need to use the information provided about the remaining moles of each gas.
Given:
- Moles of nitrogen in the container = 3.0 mol
- Moles of hydrogen in the container = 3.0 mol
- Moles of ammonia in the container = 3.0 mol
Since ammonia is formed by the reaction of nitrogen and hydrogen, we can use the balanced chemical equation for the reaction to determine the stoichiometric ratio between them.
The balanced chemical equation for the reaction is:
N2 + 3H2 -> 2NH3
From the balanced equation, we can see that 1 mole of nitrogen reacts with 3 moles of hydrogen to form 2 moles of ammonia.
Since the moles of ammonia are equal to the moles of nitrogen (3.0 mol) at the end of the reaction, we know that the initial moles of nitrogen were also 3.0 mol.
Similarly, since the moles of ammonia are equal to the moles of hydrogen (3.0 mol) at the end of the reaction, we can conclude that the initial moles of hydrogen were also 3.0 mol.
Therefore, the initial moles of both nitrogen and hydrogen were 3.0 mol each.
Let's try this.
N2 + 3H2 ==> 2NH3
We start with an unknown amount of N2 and H2 but zero NH3. If we end up with 3.0 mols NH3, we must have used up 3.0/2 or 1.5 mols N2 and 1.4*3 = 4.5 mols H2 as shown in the equation.
....N2 + 3H2 ==> 2NH3
I....x.....y........0
C..-1.5..-4.5......3
E...3.0..3.0......3.0
So x must be 1.5+3.0 = ?
y = 4.5 + 3.0 = ?
How does that look to you?