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).