A sample of 7.85 g of liquid 1‑propanol, C3H8O, is combusted with 54.0 g of oxygen gas. Carbon dioxide and water are the products.

Question

1.Chemical Reaction

2 Limiting Reaction

3 how many grams of CO2 released

4 how many grams of the excess reactant remaining

Answer 1

First, write a correctly balanced equation for the reaction taking place:

(1). 2C₃H₈O + 9O₂ ==> 6CO₂ + 8H₂O .. balanced equation

Since we are given the amounts of BOTH reactants, we must find the limiting reactant. One way to do this is to divide the MOLES of each reactant by the corresponding coefficient in the balanced equation. Whichever result is less represents the limiting reactant:

For C₃H₈O: 7.85 g x 1 mol / 60 g = 0.131 moles C₃H₈O (÷2->0.07)

For O₂: 54.0 g x 1 mol / 32 g = 1.69 moles O₂ (÷9->0.19)

(2). Since 0.07 is less than 0.19, C₃H₈O is the limiting reactant

(3). Now, use the MOLES of the limiting reactant to find the amount of product formed.

0.131 mols C₃H₈O x 6 mol CO₂ / 2 mol C₃H₈O x 44 g CO₂ / mol CO₂ = 17.3 g CO₂

(4). 0.131 mols C₃H₈O x 9 mols O₂ / 2 mols C₃H₈O = 0.590 mols O₂ USED UP

1.69 mols O₂ - 0.590 mols O₂ = 1.10 moles O₂ LEFT OVER

1.10 mols O₂ x 32 g O₂ / mol O₂ = 35.2 g O₂ LEFT OVER