The oxidation of phosphine (PH3) to phosphorus pentoxide (P2O5) is given by the chemical equation:
2PH3 + 4O2 → P2O5 + 3H2O
If the reaction gives 64% yield, how many grams of PH3 should be used to produce an actual yield of 250 grams of P2O5? (
First, we need to calculate the theoretical yield of P2O5 based on the given actual yield of 250 grams and the percent yield of 64%.
The theoretical yield can be calculated using the formula:
Theoretical yield = Actual yield / Percent yield
Theoretical yield = 250 / 0.64
Theoretical yield = 390.625 grams
Now, we can use the stoichiometry of the balanced chemical equation to determine the amount of PH3 required to produce 390.625 grams of P2O5.
From the balanced chemical equation, the molar ratio between PH3 and P2O5 is 2:1. This means that 2 moles of PH3 are required to produce 1 mole of P2O5.
First, we need to find the molar mass of P2O5:
P2O5 = (2 x 30.97 g/mol) + (5 x 16.00 g/mol)
P2O5 = 141.94 g/mol
Now we can calculate the moles of P2O5:
Moles of P2O5 = 390.625 / 141.94
Moles of P2O5 = 2.75 moles
Since the molar ratio is 2:1 for PH3 to P2O5, we need half as many moles of PH3 to produce the same amount of P2O5:
Moles of PH3 = 2.75 / 2
Moles of PH3 = 1.375 moles
Finally, we can calculate the mass of PH3 required:
Mass of PH3 = Moles of PH3 x Molar mass of PH3
Mass of PH3 = 1.375 x (1 x 30.97) // since PH3 has a molar mass of 30.97 g/mol
Mass of PH3 = 42.49 grams
Therefore, 42.49 grams of PH3 should be used to produce an actual yield of 250 grams of P2O5 with a 64% yield.