In Part A, you found the amount of product (2.60 mol P2O5) formed from the given amount of phosphorus and excess oxygen. In Part B, you found the amount of product (2.20 mol P2O5) formed from the given amount of oxygen and excess phosphorus. Now, determine how many moles of P2O5 are produced from the given amounts of phosphorus and oxygen. Express your answer to three significant figures and include the appropriate units.

Equation of the reaction:

P₄ + 5O₂ → 2P₂O₅.

Given:

Mass of P₄ = 161 g.

Molar mass = 124 g/mol

Number of moles of P₄ = mass/molar mass

n (P₄) = 161 g : 124 g/mol.

= 1.3 mol.

Mass of O₂ = 176 g.

Molar mass of O₂ = 2 * 16

= 32 g/mol

Number of moles = 176 g : 32 g/mol.

= 5.5 moles

Calculating the limiting reactant,

5.5 mol of O2/5 mole of O2 * 1 mole of P₄

= 2.75 moles of P₄ (> 1.3 moles)

P₄ is the limiting reagent.

By stoichiometry, 1 mole of P₄ gives 2 mole of P₂O₅

Therefore, number of moles of P₂O₅ = 2 * 1.3

= 2.6 moles of P₂O₅ was formed.