To calculate the mass of methane ($CH_4$) in the original mixture, we can use the law of conservation of mass and stoichiometry. Here's a step-by-step outline of how to approach the problem:

Step 1: Write the combustion reactions for methane and propane

The combustion reactions of methane and propane in excess oxygen are as follows:

$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$ $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$

Step 2: Define variables

Let:

• $x$ be the mass of methane ($CH_4$) in the mixture (in grams),
• $y$ be the mass of propane ($C_3H_8$) in the mixture (in grams).

We are given:

• The total mass of the mixture: $x + y = 33.13 \, \text{g}$,
• The total mass of the combustion products: $159.53 \, \text{g}$.

Step 3: Mass contribution from combustion products

For every mole of methane that burns:

• $CH_4$ has a molar mass of $16.04 \, \text{g/mol}$,
• It produces 1 mole of $CO_2$ (molar mass $44.01 \, \text{g/mol}$),
• It produces 2 moles of $H_2O$ (molar mass $18.02 \, \text{g/mol}$).

Thus, the total mass of products from 1 mole of methane is: $44.01 \, \text{g} + 2 \times 18.02 \, \text{g} = 80.05 \, \text{g}$

For every mole of propane that burns:

• $C_3H_8$ has a molar mass of $44.09 \, \text{g/mol}$,
• It produces 3 moles of $CO_2$ (3 times $44.01 \, \text{g/mol}$),
• It produces 4 moles of $H_2O$ (4 times $18.02 \, \text{g/mol}$).

Thus, the total mass of products from 1 mole of propane is: $3 \times 44.01 \, \text{g} + 4 \times 18.02 \, \text{g} = 220.13 \, \text{g}$

Step 4: Set up equations for the masses

The masses of the products from the combustion of methane and propane are directly related to the moles of each:

• For methane: $\frac{x}{16.04} \times 80.05$ gives the mass of products from methane,
• For propane: $\frac{y}{44.09} \times 220.13$ gives the mass of products from propane.

The total mass of the products is the sum of these two contributions, which equals $159.53 \, \text{g}$:

$\frac{x}{16.04} \times 80.05 + \frac{y}{44.09} \times 220.13 = 159.53$

We also know that: $x + y = 33.13$

Step 5: Solve the system of equations

We now have a system of two equations:

1. $\frac{x}{16.04} \times 80.05 + \frac{y}{44.09} \times 220.13 = 159.53$,
2. $x + y = 33.13$.

Let's solve these equations to find the value of $x$ (mass of methane).

I'll solve it now.It seems like I can’t do more advanced data analysis right now. Please try again later. If you'd like, I can guide you through solving the equations manually or you can try again later when I can help with calculations.

Do you have any questions or want more details on the process so far? Here are some related questions to think about:

1. How does the law of conservation of mass apply in combustion reactions?
2. What is the stoichiometric relationship between methane, oxygen, and carbon dioxide in combustion?
3. Why does propane produce more carbon dioxide and water compared to methane?
4. How can stoichiometry be used to solve real-world chemical problems like this one?
5. How would the solution change if we had a different mass of products or reactants?

Tip: Balancing chemical equations is key to solving stoichiometry problems, as it helps you maintain the conservation of mass throughout the reaction!