Consider the combustion of propane (C3H8) in the presence of oxygen:

C3H8+5O2 => 3CO2+4H2O

How many grams of O2 are required to react completely with 3 moles of propane?

Using the coefficients in the balanced equation, convert 3 moles propane to moles oxygen.

3 moles C3H8 x (5 moles O2/1 mole C3H8) = ?? moles O2.
Then convert moles O2 to grams. g = moles x molar mass.

To determine the amount of oxygen (O2) required to react completely with a given amount of propane (C3H8), you need to use the balanced chemical equation for the combustion reaction and the stoichiometry of the reaction.

The balanced equation is: C3H8 + 5O2 -> 3CO2 + 4H2O

From the balanced equation, you can see that 1 mole of propane (C3H8) requires 5 moles of oxygen (O2) to react completely.

Given that you have 3 moles of propane (C3H8), you can calculate the required amount of oxygen (O2) as follows:

Number of moles of oxygen (O2) = 5 moles of oxygen (O2) / 1 mole of propane (C3H8) * 3 moles of propane (C3H8)

Number of moles of oxygen (O2) = 15 moles of oxygen (O2)

Now, to convert the number of moles of oxygen (O2) to grams, you need to know the molar mass of oxygen, which is 32 g/mol.

Number of grams of oxygen (O2) = 15 moles of oxygen (O2) * 32 g/mol

Number of grams of oxygen (O2) = 480 g of oxygen (O2)

Therefore, 480 grams of oxygen (O2) are required to react completely with 3 moles of propane (C3H8).