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. Write the balanced chemical equation for the reaction.

Question

What is the limiting reactant?

What is the mass of CO2 released?

What is the mass excess reactant remaining?

Answer 1

First, we need a balanced equation for the reaction:

2 C3H8O + 9 O2 → 6 CO2 + 8 H2O

Next, we need to calculate the moles of each reactant from the masses provided:

7.5 g 1-propanol x (1mol 1-propanol/60.1 g 1-propanol) = 0.1247 mol 1-propanol

54 g O2 x (1mol O2/31.99g O2) = 1.687 mol O2

Then we need to identify which reactant is the limiting reactant:

0.1247 mol 1-propanol x (9 mol O2/2 mol 1-propanol) = 0.5615 mol O2 needed to fully react with 1-propanol in this reaction, which we definitely have (1.687 mol O2)

1.687 mol O2 x (2 mol 1-propanol/9 mol O2) = 0.3748 mol 1-propanol needed to fully react with O2 in this reaction, which we DO NOT have (0.1247 mol 1-propanol)

Therefore, our limiting reactant is 1-propanol, which makes sense because most combustion that occurs in nature is limited by the hydrocarbon fuel source and not the massive amounts of oxygen in the atmosphere.

Next, let’s calculate the mass of CO2 released in this reaction:

0.1247 mol 1-propanol x (6 mol CO2/2 mol 1-propanol) x (44.01 g CO2 / 1 mol CO2) = 16.46 g CO2 released

Finally, the excess reactant:

1.687 mol O2 - 0.5615 mol O2 = 1.1255 mol O2 left x (31.99 g O2/1mol O2) = 36 g O2 excess