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.
do i go from N to NH3 or H to NH3?
ammonia: 6 produced
H2: 0 remain since the 3H2 where combined
Moles of N2 remain: 1 since 4-3=1
Well, in this reaction, both nitrogen (N) and hydrogen (H) are involved in the formation of ammonia (NH3). However, since the balanced equation shows that two molecules of ammonia are produced for every molecule of nitrogen and three molecules of hydrogen, we can determine the limiting reactant and the amount of ammonia produced based on either nitrogen or hydrogen.
To find the limiting reactant, we compare the number of moles of nitrogen and moles of hydrogen available. Since you provided the number of molecules instead of moles, let's convert them first:
4 molecules of nitrogen = 4/6.022 x 10^23 moles of nitrogen
9 molecules of hydrogen = 9/6.022 x 10^23 moles of hydrogen
Now, let's convert the moles of nitrogen and hydrogen to moles of ammonia:
From nitrogen:
Using the balanced equation, for every 1 mole of nitrogen, we get 2 moles of ammonia.
So, the moles of ammonia from nitrogen = (2/1) x (4/6.022 x 10^23)
From hydrogen:
Using the balanced equation, for every 3 moles of hydrogen, we get 2 moles of ammonia.
So, the moles of ammonia from hydrogen = (2/3) x (9/6.022 x 10^23)
Next, we need to check which value is smaller - the moles of ammonia from nitrogen or the moles of ammonia from hydrogen. Whichever value is smaller represents the amount of ammonia produced.
Once you've determined the limiting reactant and calculated the amount of ammonia produced, you can easily convert it back to the number of molecules.
To determine the number of molecules of ammonia (NH3) produced after the complete reaction, you need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be produced.
To find the limiting reactant, you can compare the number of moles of nitrogen (N2) and hydrogen (H2) with the stoichiometric ratio given in the balanced equation.
The balanced equation shows that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
Using this ratio, we can calculate the number of moles of NH3 that can be produced:
- For nitrogen (N2): 4 molecules = 4/6.02 x 10^23 molecules (per mole of N2)
- For hydrogen (H2): 9 molecules = 9/6.02 x 10^23 molecules (per mole of H2)
Now, we can compare the number of moles of NH3 that can be produced from each reactant:
- Moles of NH3 from N2: (4/6.02 x 10^23 molecules N2) x (2 moles NH3/1 mole N2) = (8/6.02 x 10^23) moles NH3
- Moles of NH3 from H2: (9/6.02 x 10^23 molecules H2) x (2 moles NH3/3 moles H2) = (6/6.02 x 10^23) moles NH3
Comparing the two calculations, we can see that we obtain a smaller value using H2 as the reactant. Therefore, hydrogen (H2) is the limiting reactant.
Next, we need to determine the number of moles of NH3 produced from the limiting reactant, which is H2 in this case. According to the stoichiometric ratio, 3 moles of H2 produce 2 moles of NH3.
Moles of NH3 from limiting reactant (H2) = (6/6.02 x 10^23) moles NH3
Moles of NH3 = (6/6.02 x 10^23) moles NH3 x (2 moles NH3/3 moles H2)
Finally, we can determine the number of molecules of NH3 produced by multiplying the moles of NH3 obtained by Avogadro's number (6.02 x 10^23 molecules per mole):
Number of molecules of NH3 = (6/6.02 x 10^23) moles NH3 x (2 moles NH3/3 moles H2) x (6.02 x 10^23 molecules NH3 per mole)
By plugging in the values and performing the calculations, you can find the number of molecules of NH3 produced after the complete reaction.
To determine the number of molecules of ammonia (NH3) produced after the complete reaction, you need to first identify the limiting reactant. The limiting reactant is the reactant that is completely consumed first, thereby limiting the amount of product that can be formed.
In this balanced chemical equation:
N2 (g) + 3H2 (g) → 2NH3 (g)
The mole ratio indicates that for every 1 mole of N2, 2 moles of NH3 are produced.
Similarly, for every 3 moles of H2, 2 moles of NH3 are produced.
Now let's calculate the number of moles of N2 and H2 available based on the given information:
Number of moles of N2 = Number of molecules of N2 / Avogadro's number (6.022 x 10^23)
= 4 molecules / (6.022 x 10^23 molecules/mol)
≈ 6.645 x 10^-24 mol
Number of moles of H2 = Number of molecules of H2 / Avogadro's number
= 9 molecules / (6.022 x 10^23 molecules/mol)
≈ 1.494 x 10^-23 mol
Now, we compare the mole ratios of N2 to NH3 and H2 to NH3 to determine the limiting reactant.
For N2:
1 mole of N2 produces 2 moles of NH3
For H2:
3 moles of H2 produces 2 moles of NH3
Since the mole ratio of N2 to NH3 is 1:2, and the mole ratio of H2 to NH3 is 3:2, we can see that N2 is the limiting reactant. This means that N2 will be completely consumed before the H2, resulting in the formation of 2 moles of NH3.
Since you have 4 molecules of N2, which is equivalent to 6.645 x 10^-24 mol, and the mole ratio indicates that 1 mole of N2 produces 2 moles of NH3, you can now calculate the number of moles of NH3 produced:
Number of moles of NH3 = (Number of moles of N2) x (Mole ratio of NH3 to N2)
= (6.645 x 10^-24 mol) x (2 mol NH3 / 1 mol N2)
≈ 1.329 x 10^-23 mol
Finally, to convert the number of moles of NH3 to the number of molecules, use Avogadro's number:
Number of molecules of NH3 = Number of moles of NH3 × Avogadro's number
≈ (1.329 x 10^-23 mol) × (6.022 x 10^23 molecules/mol)
≈ 7.998 x 10^0 molecules
≈ 8 molecules
Thus, after the complete reaction, approximately 8 molecules of ammonia (NH3) will be produced.