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.

Answer 1

Not sure exactly what the question is, but regardless, you must....

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

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)

Since 0.07 is less than 0.19, C₃H₈O is the limiting reactant

Now, use the MOLES of the limiting reactant to find the amount of product formed. Not knowing the exact question, the following calculation if for moles of CO₂ formed. If need moles of H₂O or mass of product, hopefully you'll know how to do that:

0.131 mols C₃H₈O x 6 mol CO₂ / 2 mol C₃H₈O = 0.393 mols CO₂ formed