How many grams of P4O10 (292.88 g/mol) form when phelpsphorous (P4, 125.52 g/mol) reacts with 16.2 L of O2 (33.472 g/mol) ) at standard temperature and pressure

40.5 g of P₄O₁₀ are produced

Explanation:

We state the reaction:

P₄ + 5O₂ → P₄O₁₀

We do not have data from P₄ so we assume, it's the excess reactant.

We need to determine mass of oxygen and we only have volumne so we need to apply density.

Density = mass / volume, so Mass = density . volume

Denstiy of oxygen at STP is: 1.429 g/L

1.429 g/L . 16.2L = 23.15 g

We determine the moles: 23.15 g . 1mol / 33.472g = 0.692 moles

5 moles of O₂ can produce 1 mol of P₄O₁₀

Our 0.692 moles may produce (0.692 . 1)/ 5 = 0.138 moles

We determine the mass of product:

0.138 mol . 292.88 g/mol = 40.5 g

20 drops hope this