how many liters of o2 gas at STP are requires to produce 134.5g of c3h6ofrom excess c6h6 and c3h6 according to the reaction C₆H₆(l) + C₃H₆(g) + O₂(g) → C₆H₆O(s) + C₃H₆O(l)

Answer 1

C₆H₆(l) + C₃H₆(g) + O2 ==> C₆H₆O(s) + C₃H₆O(l) ... balanced equation

molar mass C3H6O = 58.08 g / mole

moles C₃H₆O produced = 134.5 g x 1 mol / 58.08 g = 2.316 mols C₃H₆O

moles O₂ needed = 2.316 mols C₃H₆O x 1 mol O₂ / 1 mol C₃H₆O = 2.316 mols O₂ needed

At STP 1 mole of any ideal gas = 22.4 L

Liters of O₂ needed = 2.316 mols x 22.4 L / mole = 51.88 L of O₂ gas needed