100 g of propane is completely reacted with 1000 g of oxygen. The mole fraction of carbon dioxide in

Q: Why do we need to find the limiting reagent? Can't I just use the mass of oxygen?

In a chemical reaction, the reactants combine in specific molar ratios. If you have an excess of one reactant (like oxygen here), it won't all react. The limiting reagent (propane) is the one that gets completely consumed and therefore determines exactly how much product can be formed.

Q: How do we know water is in the gaseous state (steam) and should be included in the total moles?

Combustion reactions release a massive amount of heat, meaning the water produced is initially in the form of steam (gas). The question asks for the mole fraction in the 'resulting mixture' without specifying a cool-down period, so we must assume the steam remains part of the gaseous mixture.

Q: What if oxygen was the limiting reagent instead?

If oxygen were the limiting reagent, you would use its moles to calculate the amount of $CO_2$ and $H_2O$ produced. You would also have to calculate the amount of unreacted propane left over and include it in your total final moles.

Q: Why do we add the moles of unreacted oxygen to the total moles?

Mole fraction is the ratio of the moles of one component to the total moles of *all* components in the mixture. Even though the leftover oxygen didn't react, it is still physically present in the container and contributes to the total number of particles in the gas mixture.

Q: I got 0.1902 for the mole fraction. Why is the answer exactly 19?

The question asks for the value of $x$ to the 'nearest integer' where the mole fraction is expressed as $x \times 10^{-2}$. Since $0.1902 = 19.02 \times 10^{-2}$, rounding $19.02$ to the nearest integer gives exactly 19.

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine we have a closed container. Inside this container, we have placed **$100 	ext{ g}$ of propane ($C_3H_8$)** and **$1000 	ext{ g}$ of oxygen ($O_2$)**. When we ignite this mixture, a combustion reaction takes place.

Before we do any math, we must write down the balanced chemical equation for the combustion of propane. Propane reacts with oxygen to form carbon dioxide and water (which will be in the form of steam due to the high heat of combustion).

$$C_3H_8(g) + 5O_2(g) ightarrow 3CO_2(g) + 4H_2O(g)$$

### The Master Equation and Moles

To figure out what happens during the reaction, we need to convert our given masses into moles. The molar mass of propane is $44 	ext{ g/mol}$ ($3 	imes 12 + 8 	imes 1$), and the molar mass of oxygen gas is $32 	ext{ g/mol}$.

$$n_{C_3H_8} = \frac{100}{44} = 2.27 	ext{ mol}$$

$$n_{O_2} = \frac{1000}{32} = 31.25 	ext{ mol}$$

Now, we face a classic stoichiometry problem: **Who is the limiting reagent?** According to our balanced equation, $1 	ext{ mol}$ of propane requires exactly $5 	ext{ mol}$ of oxygen to burn completely.

So, for $2.27 	ext{ mol}$ of propane, we would need:

$$5 	imes 2.27 = 11.35 	ext{ mol of } O_2$$

But wait! We have a massive $31.25 	ext{ mol}$ of oxygen available. Since $31.25 > 11.35$, oxygen is in excess, and **propane is our limiting reagent**. Propane will dictate how much product is formed.

### Final Calculation of the Mixture

Let's calculate the moles of everything present in the container after the reaction is complete. The propane is completely consumed, so its final moles are zero.

The moles of carbon dioxide produced will be $3$ times the moles of propane:

$$n_{CO_2} = 3 	imes 2.27 = 6.81 	ext{ mol}$$

The moles of steam produced will be $4$ times the moles of propane:

$$n_{H_2O} = 4 	imes 2.27 = 9.08 	ext{ mol}$$

And don't forget the leftover oxygen! It didn't just disappear.

$$n_{O_2(	ext{left})} = 31.25 - 11.35 = 19.90 	ext{ mol}$$

The question asks for the mole fraction of $CO_2$ in the *resulting mixture*. This means we need the total moles of all gases present.

$$n_{	ext{total}} = 6.81 + 9.08 + 19.90 = 35.79 	ext{ mol}$$

Now, finding the mole fraction of $CO_2$ is straightforward:

$$\chi_{CO_2} = \frac{n_{CO_2}}{n_{	ext{total}}} = \frac{6.81}{35.79} = 0.19$$

The problem states that the mole fraction is $x 	imes 10^{-2}$. We can rewrite $0.19$ as $19 	imes 10^{-2}$.

Comparing the two, we get our final answer:

**$x = 19$**

Enter Numerical Value:

Visualized Solution

$W_{C_{3} H_{8}} = 100 g, W_{O_{2}} = 1000 g$

$C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (g)$

$n_{C_{3} H_{8}} = 2.27 mol, n_{O_{2}} = 31.25 mol$

$Limiting Reagent Analysis$

$Moles in Final Mixture$

$χ_{C O_{2}} = \frac{6.81}{35.79} = 0.19$

$x = 19$

$Conceptual Reflection$

The Sigma Insight: Stoichiometric and Volumetric Calculations

Analyzing the Setup

The Master Equation and Moles

Final Calculation of the Mixture

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } WC3​H8​​=100 g,WO2​​=1000 g

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C3​H8​(g)+5O2​(g)→3CO2​(g)+4H2​O(g)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } nC3​H8​​=2.27 mol,nO2​​=31.25 mol

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Limiting Reagent Analysis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moles in Final Mixture

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } χCO2​​=35.796.81​=0.19

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=19

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conceptual Reflection