3 H2 (g) + N2 (g) --> 2 NH3 (g)
How many liters of ammonia gas can be produced from 10.0 L of N2 and
20.0 L of H2? Assume all gases are measured at the same temperature and pressure.
The first step would be to calculate the MR's of each substance so you could get the reacting masses so:
3H2 = (3 x 2 x 1) = 6
N2 = 14 x 2 = 28
2NH3 = (2 x 14) + (2 x 3 x 1) = 34
so in theory you need 6litres of H2 and 28litres of N2 to produce 34litres of NH3.
Is the question asking for two values or are we to assume that one of them is in excess?
The next step would be to convert to the numbers given in the question. For this I am going to assume that the hydrogen is given in excess.
so instead of 28 litres of nitrogen, I now have 10 litres. First consider 1 litre of nitrogen.
28/28 = 1 so all the other numbers need to be divided by 28.
1litre of N2 and 6/28litres of H2 make 34/28litres of NH3.
so 10 litres would be 10 times all of this.
10 litres of N2 and 10 x (6/28) litres of H2 make 10 x (34/28) litres of NH3.
10 x (34/28) = 12.1 litres of NH3
I know this is confusing, I hope this makes sense.
what if you want to calculate the number of moles ammonia gas from 3.0L of H2 and 3.0L of N2
4 l
To solve this problem, we need to use the concept of stoichiometry, which relates the amounts of reactants and products in a chemical reaction.
First, we need to determine the limiting reactant, which is the reactant that is completely consumed and determines the maximum amount of product that can be formed. This can be determined by comparing the ratios of the coefficients of the reactants in the balanced chemical equation.
According to the balanced equation, the stoichiometric ratio between N2 and NH3 is 1:2. It means that for every 1 mole of N2, 2 moles of NH3 are produced. Similarly, for H2 and NH3, the stoichiometric ratio is 3:2.
Now, let's calculate the number of moles of N2 and H2 using the ideal gas law:
PV = nRT
Where:
P = pressure (assumed to be constant)
V = volume in liters
n = number of moles
R = ideal gas constant
T = temperature
Since the volume and temperature are the same for both gases, we can ignore them in our calculations. Let's focus on determining the number of moles:
For N2:
n(N2) = V(N2) / 22.4 L/mol
For H2:
n(H2) = V(H2) / 22.4 L/mol
Given that V(N2) = 10.0 L and V(H2) = 20.0 L, we can calculate the number of moles of each gas:
n(N2) = 10.0 L / 22.4 L/mol
n(N2) ≈ 0.4464 mol
n(H2) = 20.0 L / 22.4 L/mol
n(H2) ≈ 0.8929 mol
Next, we compare the number of moles of each reactant to determine the limiting reactant. In this case, since the stoichiometric ratio for H2 to NH3 is 3:2, we need 3 moles of H2 to react completely with 1 mole of N2. Therefore, N2 is the limiting reactant since we only have 0.4464 moles, which is less than what is required.
Now that we know N2 is the limiting reactant, we can use its stoichiometric ratio with NH3 to calculate the expected amount of NH3 produced:
n(NH3) = 2 * n(N2)
n(NH3) = 2 * 0.4464 mol
n(NH3) ≈ 0.8928 mol
Finally, we can convert the moles of NH3 to liters using the ideal gas law:
V(NH3) = n(NH3) * 22.4 L/mol
V(NH3) = 0.8928 mol * 22.4 L/mol
V(NH3) ≈ 19.99 L
Therefore, approximately 19.99 liters of NH3 can be produced from the given reactant volumes.