A sample of 7.39 g of liquid 1‑propanol, C3H8O, is combusted with 42.1 g of oxygen gas. Carbon dioxide and water are the products.

Answer 1

Hello, Ella,

I'll start with an unbalanced reaction:

C₃H₈O + O₂ → CO₂ + H₂O

This was slightly challenging to balance, but I think the following works:

2C₃H₈O + 9O₂ → 6CO₂ + 8H₂O

We are given the masses for the reactants, so we can calculate the moles of each that are available:

C₃H₈O (7.39g/60.0 g/mole) = 0.123 moles

O₂ (42.1 g/32.0 g/mole) = 0.554 moles

Let's check to see if there is enogh O₂ to combust with the C₃H₈O. The balanced equation tells us the molar ratio of O₂ to C₃H₈O is 9/2, or 4.5. If we start with 0.123 moles of C₃H₈O, then we would need the following:

(0.123 moles C₃H₈O)*(4 moles O₂/9 moles C₃H₈O) = 0.554 moles O₂. We are given 1.316 moles of O₂, so we'll have more than enough. In fact, there will be 0.762 moles O₂ remaining (24.4 grams). C₃H₈O is the limiting reagent.

We find the moles, and then grams, of the products in a similar fashion. I used the molar ratios of the two products relative to the C₃H₈O:

6CO₂/2C₃H₈O or 3CO₂/C₃H₈O

8H₂O/2C₃H₈O or 4H₂O/C₃H₈O

Multiply these ratios times the moles C₃H₈O consumed to find moles of each product. Then multiply by their molar masses to find grams:

CO₂: 0.369 moles or 16.3 grams

H₂O: 0.492 moles or 8.87 grams

Bob

Answer 2

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

First, we must find which reactant, if either, is limiting.

mols C₃H₈O = 7.39 g x 1 mol/60.1 g = 0.123 mols (÷2 -> 0.0615)

mols O2 = 42.1 g x 1 mol/32 g = 1.32 mols (÷9 -> 0.147)

C₃H₈O is LIMITING

grams CO₂ formed = 0.123 mols C₃H₈O x 6 mols CO₂ / 2 mols C₃H₈O x 44 g CO₂/mol = 16.2 g CO₂ formed

grams of excess reactant (O2) left over:

mols O₂ used up = 0.123 mols C₃H₈O x 9 mols O₂ / 2 mol C₃H₈O = 0.554 mols O₂ used up

mols O₂ left over = 1.32 mols - 0.554 mols = 0.766 mols left over

grams left over = 0.766 mols O₂ x 32 g O₂/mol O₂ = 24.5 g O₂ remaining