To solve this problem, we'll break it down into three parts, as requested: identifying the limiting reagent, calculating the mass of CO₂ formed, and determining how much of the excess reactant remains.

Step 1: Balanced Reaction

The balanced reaction is: $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$ This tells us that 1 mole of propane $(C_3H_8)$ reacts with 5 moles of oxygen $(O_2)$ to produce 3 moles of carbon dioxide $(CO_2)$ and 4 moles of water $(H_2O)$.

Step 2: Calculate Moles of Reactants

We need to find the moles of each reactant to determine which one is the limiting reagent.

• Molar Mass of $C_3H_8$: $C_3H_8 = (3 \times 12.01) + (8 \times 1.008) = 44.096 \, \text{g/mol}$
• Molar Mass of $O_2$: $O_2 = 2 \times 16.00 = 32.00 \, \text{g/mol}$

Now calculate the moles of each reactant:

• Moles of $C_3H_8$: $\text{moles of } C_3H_8 = \frac{14.80 \, \text{g}}{44.096 \, \text{g/mol}} = 0.3355 \, \text{mol}$

• Moles of $O_2$: $\text{moles of } O_2 = \frac{3.44 \, \text{g}}{32.00 \, \text{g/mol}} = 0.1075 \, \text{mol}$

Step 3: Identify the Limiting Reagent

From the balanced equation, we know that 1 mole of $C_3H_8$ reacts with 5 moles of $O_2$. Let's see how much $O_2$ is needed to react with 0.3355 moles of $C_3H_8$:

$\text{Required } O_2 = 0.3355 \, \text{mol} \times 5 = 1.6775 \, \text{mol}$

However, we only have 0.1075 moles of $O_2$, which is far less than the 1.6775 moles needed. This means that $O_2$ is the limiting reagent.

Step 4: Calculate the Mass of CO₂ Produced

Since $O_2$ is the limiting reagent, we will use the moles of $O_2$ to calculate the moles and then the mass of $CO_2$ produced.

From the balanced equation, 5 moles of $O_2$ produce 3 moles of $CO_2$. So, for 0.1075 moles of $O_2$, we get: $\text{Moles of } CO_2 = \frac{3}{5} \times 0.1075 = 0.0645 \, \text{mol}$

Now, calculate the mass of $CO_2$. The molar mass of $CO_2$ is: $CO_2 = 12.01 + (2 \times 16.00) = 44.01 \, \text{g/mol}$ $\text{Mass of } CO_2 = 0.0645 \, \text{mol} \times 44.01 \, \text{g/mol} = 2.84 \, \text{g}$

Step 5: Calculate the Remaining Excess Reactant (Propane)

Now, let's determine how much $C_3H_8$ remains after the reaction. We know that 5 moles of $O_2$ react with 1 mole of $C_3H_8$, so 0.1075 moles of $O_2$ will react with: $\text{Moles of } C_3H_8 = \frac{1}{5} \times 0.1075 = 0.0215 \, \text{mol}$

Now, calculate the mass of $C_3H_8$ that reacted: $\text{Mass of } C_3H_8 \text{ reacted} = 0.0215 \, \text{mol} \times 44.096 \, \text{g/mol} = 0.948 \, \text{g}$

Initially, we had 14.80 g of $C_3H_8$. So, the mass of excess $C_3H_8$ remaining is: $\text{Mass of excess } C_3H_8 = 14.80 \, \text{g} - 0.948 \, \text{g} = 13.85 \, \text{g}$

Final Answers:

1. Limiting reagent: $O_2$
2. Grams of CO₂ formed: 2.84 g
3. Excess reactant remaining: 13.85 g of $C_3H_8$

Would you like further details or explanations on any part of this process? Here are some related questions to explore:

1. How do you calculate theoretical yield in chemical reactions?
2. What are the differences between limiting reagents and excess reagents?
3. How do you determine the molar mass of complex molecules?
4. How does stoichiometry help in real-world chemical processes?
5. Why is it important to identify the limiting reagent in a reaction?

Tip: Always make sure your units are consistent throughout calculations to avoid errors!