What is the volume of NH3 produced in the following reaction when 3.0 L of N2 reacts with 4.0 L of H2?N2(g) + 3H2(g) → 2NH3(g)
To find the volume of NH3 produced in this reaction, we need to determine which reactant is limiting and use its volume to calculate the volume of NH3 produced.
Firstly, we need to determine which reactant is limiting by calculating the amount of NH3 that could be formed from both N2 and H2.
From the balanced equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
Now, let's calculate the moles present in both reactants:
3.0 L of N2:
Using the ideal gas law (PV = nRT), we can find the number of moles of N2:
n = (PV) / RT
n = (1 atm * 3.0 L) / (0.0821 L.atm/mol.K * 300 K) = 0.122 moles
4.0 L of H2:
Similarly, for H2:
n = (1 atm * 4.0 L) / (0.0821 L.atm/mol.K * 300 K) = 0.163 moles
Now, let's determine which reactant is limiting by using the stoichiometry of the reaction:
From the mole ratios in the balanced equation, it takes 1 mole of N2 to react with 3 moles of H2. Therefore, if 3 moles of H2 react, they would require 1 mole of N2 (since the mole ratio is 1:3).
Since we have more than enough N2 (0.122 moles), H2 is the limiting reactant. Therefore, the 4.0 L of H2 will determine the amount of NH3 produced.
Calculating the moles of NH3 produced:
Using the mole ratio in the balanced equation (2 moles of NH3 for every 3 moles of H2), the moles of NH3 produced will be:
0.163 moles H2 x (2 moles NH3 / 3 moles H2) = 0.109 moles NH3
Finally, converting moles of NH3 to volume at the given conditions:
Using the ideal gas law:
V = nRT/P
V = (0.109 mol * 0.0821 L.atm/mol.K * 300 K) / 1 atm
V = 2.52 L
Therefore, 2.52 L of NH3 will be produced in the reaction.