Respuesta :
Explanation:
Nitrogen will react with hydrogen to give ammonia. The balanced reaction between them is:
N₂ + 3 H₂ ----> 2 NH₃
In our problem, 15.0 g of N₂ are reacting with 15.0 g of H₂. Before we find the percent yield we will have to determine the limiting reactant.
First, let's find the number of moles of NH₃ that will be produced by 15.0 g of N₂ (with excess H₂).
molar mass of N₂ = 2 * 14.01 g/mol
molar mass of N₂ = 28.02 g/mol
mass of N₂ = 15.0 g
moles of N₂ = 15.0 g/(28.02 g/mol)
moles of N₂ = 0.535 moles
N₂ + 3 H₂ ----> 2 NH₃
According to the coefficients of the reaction, 1 mol of N₂ will produce 2 moles of NH₃. The molar ratio between N₂ and NH₃ is 1 to 2. We can use that ratio to find the number of moles of NH₃ that will be produced by 0.535 moles of N₂ (or 15.0 g of it).
1 mol of N₂ = 2 moles of NH₃
moles of NH₃ = 0.535 moles of N₂ * 2 moles of NH₃/(1 mol of N₂)
moles of NH₃ = 1.07 moles
So we found that 15.0 g of N₂ will yield 1.07 moles of NH₃. Let's do something similar to find the number of moles of NH₃ that will be produced by 15.0 g of H₂.
molar mass of H₂ = 2 * 1.01 g/mol
molar mass of H₂ = 2.02 g/mol
mass of H₂ = 15.0 g
moles of H₂ = 15.0 g/(2.02 g/mol)
moles of H₂ = 7.43 moles
3 moles of H₂ = 2 moles of NH₃
moles of NH₃ = 7.43 moles of H₂ * 2 moles of NH₃/(3 moles of H₂)
moles of NH₃ = 4.95 moles
So, 15.0 g of H₂ (reacting with excess N₂) will produce 4.95 moles of NH₃. And 15.0 g of N₂ will yield 1.07 moles of NH₃. So the reactant that is limiting our reaction is N₂ and H₂ is in excess.
N₂ = limiting reactant H₂ = excess
Now that we know the number of moles of NH₃ we can convert them into grams using its molar mass and finally find the percent yield of the reaction.
molar mass of NH₃ = 1 * 14.01 g/mol + 3 * 1.01 g/mol
molar mass of NH₃ = 17.04 g/mol
mass of NH₃ = 1.07 moles * 17.04 g/mol
mass of NH₃ = 18.2 g = theoretical yield
actual yield of NH₃ = 10.5 g
% yield = actual yield/theoretical yield * 100
% yield = 10.5 g/18.2 g * 100
% yield = 57.7 %
Answer: the percent yield for the reaction is 57.7 %.
