Question

What volume of oxygen at STP is needed for the complete combustion of 3,3 dm^3 of propane (C3H8)
(hint :CO2 and H2O are the products as all combustion reaction)
C3H8 +5O2 (arrow) 3CO2 +4H2O

Answer 1

To determine the volume of oxygen needed for the complete combustion of 3.3 dm³ of propane (C₃H₈) at standard temperature and pressure (STP), we can use the stoichiometry of the combustion reaction.

The balanced combustion reaction for propane is:

\[
\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}
\]

From this equation, we can see that 1 mole of propane reacts with 5 moles of oxygen. At STP, 1 mole of any gas occupies 22.4 dm³.

### Step 1: Calculate the number of moles of propane.

Since we are given the volume of propane (3.3 dm³), we can calculate the number of moles of propane:

\[
\text{Number of moles of } C_3H_8 = \frac{3.3 \, \text{dm}^3}{22.4 \, \text{dm}^3/\text{mol}} \approx 0.1473 \, \text{mol}
\]

### Step 2: Calculate the number of moles of oxygen needed.

According to the balanced equation, for every 1 mole of propane, 5 moles of oxygen are required. Therefore:

\[
\text{Moles of } O_2 = 5 \times \text{moles of } C_3H_8 = 5 \times 0.1473 \, \text{mol} \approx 0.7365 \, \text{mol}
\]

### Step 3: Calculate the volume of oxygen required.

Now, we can find the volume of oxygen needed at STP:

\[
\text{Volume of } O_2 = 0.7365 \, \text{mol} \times 22.4 \, \text{dm}^3/\text{mol} \approx 16.52 \, \text{dm}^3
\]

### Conclusion

The volume of oxygen needed for the complete combustion of 3.3 dm³ of propane at STP is approximately **16.52 dm³**.