chem homework help

Question

A sample of 6.35 g6.35 g of liquid 1‑propanol, C3H8O,C3H8O, is combusted with 43.7 g43.7 g of oxygen gas. Carbon dioxide and water are the products.

Write the balanced chemical equation for the reaction. Physical states are optional.

What is the limiting reactant?

How many grams of CO2 are released in the reaction?

How many grams of the excess reactant remain after the reaction is complete?

Answer 1

C₃H₈ + 5O₂ ==> 3CO₂ + 4H₂O ... balanced equation

Find the limiting reactant:

moles C₃H₈ = 6.35 g x 1 mol / 44 g = 0.144 mols

moles O₂ = 43.7 g x 1 mol / 32 g = 1.37 mols

C₃H₈ is limiting since it takes 5 mol O₂ per mol C₃H₈ and we have 9.5 mols O₂ per mol.

Grams CO₂ released = 0.144 mol C₃H₈ x 3 mol CO₂ / mol C₃H₈ x 44 g CO₂/mol = 19.0 g CO₂

Excess reactant (O₂) = initial amount - amount used and amount used = 0.144 mol C₃H₈ x 5 mol O₂/mol = 0.72 mol O₂ used

mols O₂ in excess = 1.37 mol - 0.72 mol = 0.65 mol O₂ in excess

grams O₂ in excess = 0.65 mol O₂ x 32 g / mol = 20.8 g O₂ in excess