Answer:
a) The limiting reactant is O2.
b) 7.57 grams of P4O10 is produced
c) 7.53 grams P4O6 remains
Explanation:
Step 1: Data given
Mass of P4 = 7.55 grams
Mass of O2 = 7.55 grams
Molar mass of P4 = 123.90 g/mol
Molar mass of O2 = 32 g/mol
Step 2: The balanced equations:
P4 + 3O2-→P4O6
P4O6 + 2O2 → P4O10
Step 3: Calculate moles P4
Moles P4 = mass P4 / molar mass P4
Moles P4 = 7.55 grams / 123.90 g/mol
Moles P4 = 0.0609 moles
Step 4: Calculate moles O2
Moles O2 = 7.55 grams / 32.0 g/mol
Moles O2 = 0.236 moles
Step 5: Calculate the limiting reactant
For 1 mol P4 we need 3 moles O2 to produce 1 mol P4O6
P4 is the limiting reactant. It will completely be consumed. (0.0609 moles)
O2 is in excess. There will react 3*0.0609 = 0.1827 moles
There will remain 0.236 - 0.1827 = 0.0533 moles O2
Step 6: Calculate moles P4O6
For 1 mol P4 we need 3 moles O2 to produce 1 mol P4O6
For 0.0609 moles P4 we will have 0.0609 moles P4O6
Step 7: Calculate limting reactant
There remain 0.0533 moles O2 and there are 0.0609 moles P4O6 produced
For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O6
<u>The limiting reactant is O2.</u> It will completely be reacted (0.0533 moles)
There will react 0.0533/2 = 0.02665 moles
There will remain 0.0609 - 0.02665 = 0.03425 moles P4O6
This is 0.03425 moles * 219.88 g/mol = <u>7.53 grams P4O6</u>
Step 8: Calculate moles P4O10
For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O6
For 0.0533 moles O2, we'll have 0.0533/2 = 0.02665 moles P4O10
Step 9: Calculate mass P4O10
Mass P4O10 = 0.02665 moles * 283.89 g/mol
Mass P4O10 = <u>7.57 grams</u>
