Question

The oxidation of phosphine (PH3) to phosphorus pentoxide (P2O5) is given by the chemical equation:

2PH3 + 4O2 → P2O5 + 3H2O



How many grams of P2O5 and H2O will be produced in the reaction?

Answer 1

To calculate the amount of P2O5 and H2O produced in the reaction, we first need to determine the molar masses of each compound:

P2O5:
1 mol of P2O5 contains 2 mol of P and 5 mol of O.
Molar mass of P: 30.97 g/mol
Molar mass of O: 16.00 g/mol
Molar mass of P2O5 = 2(30.97) + 5(16.00) = 141.94 g/mol

H2O:
Molar mass of H: 1.01 g/mol
Molar mass of O: 16.00 g/mol
Molar mass of H2O = 2(1.01) + 16.00 = 18.02 g/mol

Now we need to calculate the number of moles of PH3 present in the reaction. Since the stoichiometry of the reaction is 2 mol PH3 to 1 mol P2O5, we can use the mole ratio to find the number of moles of P2O5 and H2O produced.

Given that the molar mass of PH3 is 33.997 g/mol, we use this and the mass of PH3 to find the number of moles present:

molar mass of PH3 = 33.997 g/mol
mass of PH3 = 50.0 g

Number of moles of PH3 = mass/molar mass
Number of moles of PH3 = 50.0 g / 33.997 g/mol
Number of moles of PH3 = 1.4716 mol

Now, using the mole ratio from the balanced equation, we can determine the number of moles of P2O5 and H2O produced:

1.4716 mol PH3 x (1 mol P2O5 / 2 mol PH3) = 0.7358 mol P2O5
1.4716 mol PH3 x (3 mol H2O / 2 mol PH3) = 2.2074 mol H2O

Finally, we can calculate the mass of P2O5 and H2O produced:

Mass of P2O5 = number of moles x molar mass
Mass of P2O5 = 0.7358 mol x 141.94 g/mol = 104.62 g

Mass of H2O = number of moles x molar mass
Mass of H2O = 2.2074 mol x 18.02 g/mol = 39.77 g

Therefore, 104.62 grams of P2O5 and 39.77 grams of H2O will be produced in the reaction.