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

explanation: it gains negative electrons, therefore becoming negatively charged. as charge cannot be created, nor destroyed, only transferred, this leaves the wool positive. now that they have opposite charges, the two items attract.