Question

100 g of propane is completely reacted with 1000 g of oxygen. The mole fraction of carbon dioxide in the resulting mixture is \(x \times 10^{-2}\). The value of x is _________. (Nearest integer)
[Atomic weight : H=1.008; C=12.00; O=16.00]

Hint:

In stoichiometry problems, follow these steps systematically: 1. Write a balanced equation. 2. Convert all given masses to moles. 3. Determine the limiting reactant. 4. Calculate the moles of products and excess reactants based on the limiting reactant. 5. Use the mole values to find the required quantity (e.g., mole fraction, mass).

Answer 1

Correct Answer: 19

Step 1: Understanding the Question:
This is a limiting reactant stoichiometry problem. We need to find the composition of the product mixture after a complete reaction and then calculate the mole fraction of carbon dioxide.
Step 2: Balanced Chemical Equation:
The complete combustion of propane (C\(_3\)H\(_8\)) is:
\[ \text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g) \] Step 3: Calculate Moles of Reactants:
- Molar mass of propane (C\(_3\)H\(_8\)) = 3(12.00) + 8(1.008) = 36.00 + 8.064 = 44.064 g/mol.
Moles of C\(_3\)H\(_8\) = \( \frac{100 \, \text{g}}{44.064 \, \text{g/mol}} \approx 2.269 \) mol.
- Molar mass of oxygen (O\(_2\)) = 2(16.00) = 32.00 g/mol.
Moles of O\(_2\) = \( \frac{1000 \, \text{g}}{32.00 \, \text{g/mol}} = 31.25 \) mol.
Step 4: Identify the Limiting Reactant:
From the balanced equation, 1 mole of C\(_3\)H\(_8\) requires 5 moles of O\(_2\).
- Moles of O\(_2\) required to react with 2.269 mol of C\(_3\)H\(_8\) = \( 2.269 \times 5 = 11.345 \) mol.
- Since we have 31.25 mol of O\(_2\), which is more than 11.345 mol, oxygen is in excess and propane is the limiting reactant.
Step 5: Calculate Moles in the Final Mixture:
The reaction goes to completion based on the limiting reactant (propane).
- Moles of CO\(_2\) produced = \( 3 \times \text{moles of C}_3\text{H}_8 = 3 \times 2.269 = 6.807 \) mol.
- Moles of H\(_2\)O produced = \( 4 \times \text{moles of C}_3\text{H}_8 = 4 \times 2.269 = 9.076 \) mol.
- Moles of O\(_2\) remaining = \( \text{Initial moles} - \text{Reacted moles} = 31.25 - 11.345 = 19.905 \) mol.
- Total moles in the final mixture = Moles(CO\(_2\)) + Moles(H\(_2\)O) + Moles(O\(_2\))
Total moles = \( 6.807 + 9.076 + 19.905 = 35.788 \) mol.
Step 6: Calculate Mole Fraction of CO\(_2\):
Mole fraction \( X_{\text{CO}_2} = \frac{\text{moles of CO}_2}{\text{total moles}} = \frac{6.807}{35.788} \approx 0.1902 \)
Step 7: Find the value of x:
We are given that the mole fraction is \( x \times 10^{-2} \).
\[ 0.1902 = x \times 10^{-2} \] \[ x = 19.02 \] The value of x to the nearest integer is 19.