Use the coefficients to tell you.
13.7 mols N2 x (4 mols NH3/5 mols N2) = ? mols NH3
13.7 mols N2 x (6 mols NO/5 mols N2) = ? mols NO
4NH3(g) + 6NO(g)→5N2(g) + 6H2O(g)
13.7 mols N2 x (4 mols NH3/5 mols N2) = ? mols NH3
13.7 mols N2 x (6 mols NO/5 mols N2) = ? mols NO
From the given balanced equation:
4NH3(g) + 6NO(g) → 5N2(g) + 6H2O(g)
We can see that the ratio of NH3 to N2 is 4:5.
Given that 13.7 moles of N2(g) is produced, we can set up a proportion using the ratio of coefficients:
4 moles of NH3 corresponds to 5 moles of N2
x moles of NH3 corresponds to 13.7 moles of N2
Using this proportion, we can solve for x (the number of moles of NH3):
(4 moles of NH3 / 5 moles of N2) = (x moles of NH3 / 13.7 moles of N2)
Cross-multiplying:
4 * 13.7 = 5 * x
54.8 = 5x
Dividing both sides by 5:
54.8 / 5 = x
x ≈ 10.96
Therefore, there were approximately 10.96 moles of NH3.
To find the number of moles of NO, we can use the same approach:
6 moles of NO corresponds to 5 moles of N2
y moles of NO corresponds to 13.7 moles of N2
Using the proportion:
(6 moles of NO / 5 moles of N2) = (y moles of NO / 13.7 moles of N2)
Cross-multiplying:
6 * 13.7 = 5 * y
82.2 = 5y
Dividing both sides by 5:
82.2 / 5 = y
y ≈ 16.44
Therefore, there were approximately 16.44 moles of NO.
From the balanced equation:
4NH3(g) + 6NO(g) -> 5N2(g) + 6H2O(g)
The coefficient of NH3 is 4, and the coefficient of N2 is 5. This means that for every 4 moles of NH3, 5 moles of N2 are produced.
Since we know that 13.7 moles of N2(g) is produced, we can set up a ratio to find the moles of NH3:
(13.7 moles N2) x (4 moles NH3 / 5 moles N2) = 10.96 moles NH3
Therefore, there were approximately 10.96 moles of NH3 present.
Similarly, we can find the moles of NO by setting up a ratio:
(13.7 moles N2) x (6 moles NO / 5 moles N2) = 16.44 moles NO
Therefore, there were approximately 16.44 moles of NO present.
So, the number of moles of each reactant were approximately 10.96 moles of NH3 and 16.44 moles of NO.