Answer:
a) O2 is the limiting reactant
b) 5.75 grams P4O10
c) 5.79 grams P4O6
Explanation:
Step 1: Data given
Mass of P4 = 5.77 grams
Mass of O2 = 5.77 grams
Molar mass of P4 = 123.90 g/mol
Molar mass O2 = 32.0 g/mol
Step 2: The balanced equation
P4 + 3O2 → P4O6
Step 3: Calculate moles of P4
Moles P4 = mass P4 / molar mass P4
Moles P4 = 5.77 grams / 123.90 g/mol
Moles P4 = 0.0466 moles
Step 4: Calculate moles O2
Moles O2 = mass O2 / molar mass O2
Moles O2 = 5.77 grams / 32.0 g/mol
Moles O2 = 0.1803 moles
Step 5: Calculate limiting reactant
P4 is the limiting reactant in this reaction. It will completely be consumed (0.0466 moles). O2 is in excess, there will react 3*0.0466 = 0.1398 moles
There will remain 0.1803 - 0.1398 = 0.0405 moles O2
Step 6: Calculate the amount of P4O6
For 1 mol P4 we'll have 1 mol P4O6
For 0.0466 moles P4 we'll have 0.0466 moles P4O6
Step 7: The balanced equatio
P4O6 + 2O2 → P4O10
We have 0.0466 moles P4O6 and 0.0405 moles O2
Step 8: Calculate the limiting reactant
For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O10
O2 is the limiting reactant. It will completely be consumed (0.0405 moles)
P4O6 is in excess. There will react 0.0405/2 = 0.02025 moles
There will remain 0.0466 - 0.02025 = 0.02635 moles P4O6
This is 0.02635 * 219.88 g/mol = 5.79 grams P4O6
Step 9: Calculate moles and mass of P4O10
For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O10
For 0.0405 moles O2 we'll have 0.02025 moles P4O10
This is 0.02025 * 283.89 g/mol = 5.75 grams P4O10
