Chemistry Stoichiometry calculations

Question

If 45.0g of C3H8 react with 45.0g of O2 what is the limiting reactant? How much CO2 is produced?

Answer 1

To determine the limiting reactant, we need to compare the number of moles of C3H8 and O2 available for the reaction. We can use the balanced chemical equation for the combustion of propane:

C3H8 + 5O2 → 3CO2 + 4H2O

From the equation, we can see that one mole of C3H8 reacts with five moles of O2 to produce three moles of CO2. Therefore, we need to convert the masses of C3H8 and O2 to moles:

n(C3H8) = 45.0 g / 44.1 g/mol = 1.02 mol

n(O2) = 45.0 g / 32.0 g/mol = 1.41 mol

Now we can use the mole ratios from the balanced equation to determine which reactant is limiting:

1. For C3H8, the expected amount of CO2 produced is 3 moles of CO2 per 1 mole of C3H8, or 3.06 moles of CO2 for 1.02 moles of C3H8.
2. For O2, the expected amount of CO2 produced is 3 moles of CO2 per 5 moles of O2, or 0.84 moles of CO2 for 1.41 moles of O2.

Since the expected amount of CO2 produced is less for O2 than for C3H8, O2 is the limiting reactant. This means that all of the O2 will be consumed in the reaction, and there will be some C3H8 left over.

To calculate the amount of CO2 produced, we can use the mole ratio from the balanced equation between O2 and CO2:

1. For every 5 moles of O2, 3 moles of CO2 are produced.
2. Therefore, for 1.41 moles of O2, the expected amount of CO2 produced is:
3. (3/5) * 1.41 mol = 0.84 mol

Therefore, the limiting reactant is O2 and the amount of CO2 produced is 0.84 moles, or:

0.84 mol CO2 * 44.01 g/mol = 37.0 g CO2.

So, 37.0 g of CO2 is produced.