Question

how many grams of phosphorus (P4) react with 35.5L of O2 at STP to form solid tetraphosphorus decaoxide

Answer 1

Answer:

mass P4 = 35.998 g

Explanation:

• P4 + 5O2 → P4O10

∴ STP: P = 1 atm; T = 298 K

∴ V O2= 35.5 L

⇒ nO2 = P.V / R.T

∴ R = 0.082 atm.L/K.mol

⇒ nO2 = ((1 atm)×(35.5L))/((0.082 atm.L/K.mol)(298K))

⇒ nO2 = 1.453 mol O2

⇒ mol P4 = (1.453 molO2)×(mol P4/ 5molO2) = 0.2906 mol P4

∴ Mw P4 = 123.895 g/mol

⇒ mass P4 = (0.2906 mol P4)×(123.895 g/mol) = 35.998 g P4