PLEASE help with chem limiting reactants

Question

2 C3H8 (g) + 10 O2 (g) --- 6 CO2 (g) + 8 H2O (g)

Given there are 15 grams of propane gas (C3H8) and 40 grams of o2

a.) Which is the limiting Reagent?

B: how many grams of carbon dioxide are produced

Answer 1

2 C₃H₈ (g) + 10 O₂ (g) --- 6 CO₂ (g) + 8 H₂O (g) ... balanced equation

a). Divide the moles of each reactant by the corresponding coefficient in the balanced equation. Whichever values comes out less is the limiting reactant.

For C₃H₈ we have .. 15 g C₃H₈ x 1 mol / 44.1 g = 0.340 mols C₃H₈ (÷2->0.17)

For O₂ we have .. 40 g O₂ x 1 mol / 32 g = 1.25 mols O₂ (÷10->0.125)

Since 0.125 is less than 0.17, O₂ is the limiting reactant

b). Use the moles of the limiting reactant and the stoichiometry to find grams of CO₂ produced:

1.25 mols O₂ x 6 mols CO₂ / 10 mols O₂ x 44 g CO₂ / mol CO₂ = 33 g CO₂ produced