7.55 grams of P4 and 7.55 grams of O2 react according to the following reaction:

P4 + O2--> P4O6

If enough oxygen is available, then the P4O6 reacts further:

P4O6 + O2 - -> P4O10

a. Find the limiting reagent in the formation of P4O10.

b. What mass of P4O10 is produced?

c. What mass of excess reactant remains?

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

The limiting reactant is O2. 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 = 7.53 grams P4O6

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 = 7.57 grams