Question

propane combusts with oxygen according to the reaction:c3h8 5 o2 --> 3 co2 4 h2oif 15.0 g of c3h8 and 50.0g of o2 are allowed to react, which is the limiting reactant?

Answer 1

To determine the limiting reactant between propane (C3H8) and oxygen (O2), convert the masses to moles, then use the stoichiometric mole ratios from the balanced equation to compare the available moles of each reactant. Oxygen is the limiting reactant because there are insufficient moles of it to completely react with the propane.

Step-by-step explanation:

To determine the limiting reactant when propane (C3H8) combusts with oxygen (O2), we use the balanced chemical equation C3H8 + 5 O2 → 3 CO2 + 4 H2O.

First, we convert the masses of reactants to moles:

• For C3H8: Molecular weight = 44.10 g/mol; Moles of C3H8 = 15.0 g / 44.10 g/mol = 0.340 moles
• For O2: Molecular weight = 32.00 g/mol; Moles of O2 = 50.0 g / 32.00 g/mol = 1.5625 moles

Now we compare the mole ratio from the balanced equation to the actual mole ratio:

• According to the balanced equation, 1 mole of C3H8 requires 5 moles of O2.
• Thus, 0.340 moles of C3H8 would require 0.340 moles x 5 = 1.70 moles of O2 for complete combustion.
• Since we only have 1.5625 moles of O2, which is less than the required 1.70 moles, O2 is the limiting reactant.