Quantum chemistry question

The percent yield of the reaction is 48.9 %

Explanation:

Step 1: data given

Mass of phosphorus = 75.3 grams

Molar mass of phosphorus = 123.90 g/mol

Mass of oxygen = 38.7 grams

Molar mass oxygen = 32.0 g/mol

Mass of P4O6 produced = 43.3 grams

Step 2: The balanced equation

P4(s) + 3O2(g) → P4O6(s)

Step 3: Calculate moles P4

Moles P4 = mass P4 / molar mass P4

Moles P4 = 75.3 grams / 123.90 g/mol

Moles P4 = 0.608 moles

Step 4: calculate moles O2

Moles O2 = 38.7 grams / 32.0 g/mol

Moles O2 = 1.21 moles

Step 5: Calculate the limiting reactant

For 1 mol P4 we need 3 moles O2 to produce 1 mol P4O6

O2 is the limiting reactant. There will react 1.21 moles. P4 is in excess. There will react 1.21 / 3 = 0.403 moles. There will remain 0.608 - 0.403 = 0.205 moles P4

Step 6: Calculate moles P4O6

For 1 mol P4 we need 3 moles O2 to produce 1 mol P4O6

For 1.21 moles O2 we'll have 1.21 / 3 = 0.403 moles P4O6

Step 7: Calculate mass P4O6

Mass P4O6 = moles P4O6 * molar mass P4O6

Mass P4O6 = 0.403 moles * 219.88 g/mol

Mass P4O6 = 88.6 grams

Step 8: Calculate percent yield

% yield = (actual yield / theoretical yield) * 100%

% yield = (43.3 grams / 88.6 grams ) *100 %

% yield =  48.9 %

The percent yield of the reaction is 48.9 %