Final answer:
To combust 8.50 moles of propane completely, 42.5 moles of oxygen are needed. With only 36.0 moles of oxygen available, oxygen is the limiting reactant with no excess. There are 1.30 moles of propane remaining as the excess reactant.
Step-by-step explanation:
The question pertains to a chemical reaction where propane gas (C3H8) is combusted with oxygen gas (O2) to produce carbon dioxide (CO2) and water (H2O). The balanced chemical equation for the combustion of propane gas is:
C3H8 + 5 O2 → 3 CO2 + 4 H2O
According to this equation, to completely combust one mole of propane, five moles of oxygen are required. Therefore, to burn 8.50 moles of propane, 42.5 moles of oxygen are needed (8.50 moles C3H8 x 5 moles O2/mol C3H8). Since only 36.0 moles of O2 are available, oxygen is the limiting reactant, and there will be no excess oxygen remaining after the reaction.
Since propane is in excess (8.50 mol propane available, but only 7.2 mol of propane can react with 36.0 mol of oxygen, given the 1:5 ratio), we can calculate the amount of excess propane. Subtracting the amount of propane that reacts from the initial amount gives us the excess amount: 8.50 mol - (36.0 mol O2 x 1 mol C3H8 / 5 mol O2) = 8.50 mol - 7.2 mol = 1.30 moles of C3H8 remaining as excess reactant.