Answer:
Approximately 142.55 grams of carbon dioxide is produced upon the complete combustion of 24.2 liters of propane.
Explanation:
To solve this problem, we need to first determine the moles of propane [tex](\(C_3H_8\))[/tex] using its molar volume at standard conditions (STP), which is 22.4 liters per mole. Then, we use the balanced chemical equation for the combustion of propane to find the mole ratio between propane and carbon dioxide. Finally, we convert the moles of carbon dioxide to grams using its molar mass.
The balanced chemical equation for the complete combustion of propane [tex](\(C_3H_8\))[/tex]
is:
[tex]\[ C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g) \][/tex]
From the equation, we see that 1 mole of propane produces 3 moles of carbon dioxide.
Given that the volume of propane is 24.2 L, we can calculate the number of moles of propane:
[tex]\[ \text{Number of moles of } C_3H_8 = \frac{\text{Volume of propane}}{\text{Molar volume at STP}} = \frac{24.2 \text{ L}}{22.4 \text{ L/mol}} \][/tex]
[tex]\[ \text{Number of moles of } C_3H_8 = 1.08036 \text{ moles} \][/tex]
Now, we can use the mole ratio from the balanced equation to find the moles of carbon dioxide produced:
[tex]\[ \text{Moles of } CO_2 = \text{Moles of } C_3H_8 \times \frac{3 \text{ moles of } CO_2}{1 \text{ mole of } C_3H_8} \][/tex]
[tex]\[ \text{Moles of } CO_2 = 1.08036 \times 3 \][/tex]
[tex]\[ \text{Moles of } CO_2 = 3.24108 \text{ moles} \][/tex]
Finally, we convert moles of carbon dioxide to grams using its molar mass:
[tex]\[ \text{Mass of } CO_2 = \text{Moles of } CO_2 \times \text{Molar mass of } CO_2 \][/tex]
[tex]\[ \text{Mass of } CO_2 = 3.24108 \text{ moles} \times 44.01 \text{ g/mol} \][/tex]
[tex]\[ \text{Mass of } CO_2 \approx 142.55 \text{ grams} \][/tex]
Therefore, approximately 142.55 grams of carbon dioxide is produced upon the complete combustion of 24.2 liters of propane.
