2N2o5->4no2+of2,when 40g of N2o5 and 4.5 of oxygen. Calculate theoretical yield.
To calculate the theoretical yield, we need to determine the limiting reactant first.
From the balanced equation 2N2O5 -> 4NO2 + O2, we can see that the molar ratio between N2O5 and NO2 is 2:4 or 1:2.
1 mole of N2O5 produces 2 moles of NO2.
First, we need to calculate the moles of N2O5 in 40g.
Molar mass of N2O5 = (14.01 x 2) + 15.999 x 5) = 108.01 g/mol
Moles of N2O5 = Mass / Molar mass = 40g / 108.01 g/mol ≈ 0.37 mol
Since the molar ratio is 1:2, the moles of NO2 produced will be twice the moles of N2O5. Therefore:
Moles of NO2 = 0.37 mol × 2 = 0.74 mol
Now, let's calculate the moles of O2 in 4.5g.
Molar mass of O2 = (15.999 x 2) = 31.998 g/mol
Moles of O2 = Mass / Molar mass = 4.5g / 31.998 g/mol ≈ 0.14 mol
Since the molar ratio between N2O5 and O2 is 1:1, the moles of O2 required to react completely with the 0.37 mol of N2O5 would also be 0.37 mol.
However, we only have 0.14 mol of O2, which is less than what is required. Therefore, O2 is the limiting reactant.
The moles of NO2 produced will be half of the moles of O2.
Moles of NO2 = 0.14 mol / 2 = 0.07 mol
Now, let's calculate the mass of NO2 produced.
Molar mass of NO2 = (14.01 + 15.999 x 2) = 46.01 g/mol
Mass of NO2 = Moles × Molar mass = 0.07 mol × 46.01 g/mol = 3.22 g
Therefore, the theoretical yield of NO2 is 3.22 grams.