The chemical equation below shows the combustion of propane (c3h8). c3h8 + 5o2 mc012-1.jpg 3co2 + 4h2o the molar mass of oxygen gas (o2) is 32.00 g/mol. the molar mass of c3h8 is 44.1 g/mol. what mass of o2, in grams, is required to completely react with 0.025 g c3h8?

The balanced equation for the reaction is 
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Moles (mol) = mass (g) / molar mass (g/mol)

Mass of the C₃H₈ = 0.025 g
Molar mass of C₃H₈ = 44.1 g/mol
Hence, moles of C₃H₈  = 0.025 g / 44.1 g/mol = 5.67 x 10⁻⁴ mol

The stoichiometric ratio between C₃H₈ and O₂ is 1 : 5.

Hence, moles of O₂ = moles of C₃H₈ x 5
                                = 5.67 x 10⁻⁴ mol x 5
                                = 2.835 x 10⁻³ mol

Molar mass of O₂ = 32.00 g/mol
Hence, mass of O₂ = moles x molar mass
                               = 2.835 x 10⁻³ mol x 32.00 g/mol 
                               = 0.09072 g

Hence, needed O₂ for the reaction is 0.09072 g.

Ithink this is what you meant. the temperature at which a liquid begins to boil is called the boiling point. the boiling point of water is 212°f or 100°c.  and at that point will start to turn into steam (gas)