A reaction container holds 5.77 g of P4 and 5.77 g of O2.
The following reaction occurs: P4 + O2 → P4O6.
If enough oxygen is available then the P4O6 reacts further: P4O6 + O2 → P4O10.
a. What is the limiting reagent for the formation of P4O10?
b. What mass of P4O10 is produced?
c. What mass of excess reactant is left in the reaction container?

a. The limiting reagent for the formation of P₄O₁₀ is O₂

b. The mass of P₄O₁₀ produced is 5.79 g

c. The mass of excess reactant left is 5.76138  g  

Firstly, the limiting reagent is the reactant that is completely used up in a reaction while the excess reagent is the reactant that is not used up when the reaction is concluded.

Let's represents the series of the reactions chemically and balance them.

P₄ + O₂→ P₄O₆

P₄ + 3O₂→ P₄O₆

P₄O₆ + O₂→P₄O₁₀

P₄O₆ +2O₂→P₄O₁₀

Let's calculate the values to get the moles/mass of P₄O₆ that was used in the final chemical equation.

Therefore,

Moles of P₄ = mass/molar mass  =5.77/123.90 = 0.04656981436≈0.0465 moles

Moles of O₂ = 5.77/32 = 0.1803125≈0.1803 moles

From  the first chemical reaction, P₄  is the limiting reagent while O₂ is the excess reactant.

This means 0.0465 moles of P₄ reacted with 3 × 0.0465 moles of O₂(0.1395

moles)

The excess oxygen remaining in the first reaction will be 0.1803 - 0.1395 = 0.0408  moles

P₄O₆ +2O₂→P₄O₁₀

1 mole of P₄O₆ + 2 moles of O₂ = 1 mole of P₄O₁₀

0.0408 moles of O₂ will produce 0.0204  moles of P₄O₁₀

a.

The limiting reagent in the formation of P₄O₁₀ is O₂(0.0408 moles).

b.

The mass of of P₄O₁₀ produced can be below

moles = mass/molar mass =

mass = moles × molar mass

mass = 0.0204  × 283.90 = 5.79 grams

c.

The excess reactant(P₄O₆) mass will be

1 mole of P₄O₆ needs 2 moles of O₂ to produce 1 mole of P₄O₁₀

Therefore,

moles of P₄O₆  that will react will be = 0.0408/2 = 0.0204  moles

The remainder will be = 0.0466 - 0.0204  = 0.0262  moles

mass of the excess reactant left = 0.0262  × 219.9(moles × molar mass of P₄O₆) = 5.76138  g left.  

Refer to this link for more information:

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