How many grams of propane can combust at 25°C and 1.04 atm if reacting with 14.7 L of oxygen gas?

The volume of oxygen is 14.7 litres.What is volume?volume is a measure of occupied three dimensional space.Sol-Balanced chemical reaction: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.m(C₃H₈-propane) = 5.53 g.n(C₃H₈) = m(C₃H₈) ÷ M(C₃H₈).n(C₃H₈) = 5.53 g ÷ 44.1 g/mol.n(C₃H₈) = 0.125 mol.From chemical reaction: n(C₃H₈) : n(O₂) = 1 : 5.n(O₂) = 0.625 mol.T = 25° = 298.15K.p = 1.04 atm.R = 0.08206 L·atm/mol·K.Ideal gas law: p·V = n·R·T .V(O₂) = n·R·T / p.V(O₂) = 0.625 mol · 0.08206 L·atm/mol·K · 298.15 K / 1.04 atm.V(O₂) = 14.7 L.To know more about volume click-https://brainly.com/question/23963432#SPJ11