How many grams of phosphorus (p4) react with 35.5l of o2 at stp to form solid tetraphosphorus decaoxide

mass P4 = 35.998 g

Explanation:

∴ 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

water at room temperature is flowing through a pipe with pressure taps located 10 ft apart along one section of the pipe, which are connected to a mercury (s=13.6) manometer.

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