Calculating CO2 Production From Methane Combustion A Stoichiometry Guide

Hey guys! Let's dive into a super important topic in chemistry: stoichiometry. Stoichiometry, in simple terms, is like the recipe book of chemical reactions. It tells us exactly how much of each ingredient (or reactant) we need to make a specific amount of the final product. Today, we're going to tackle a classic stoichiometry problem involving the combustion of methane (CH₄), the main component of natural gas. This is a reaction you see every time you light your gas stove or a Bunsen burner in the lab, so understanding it is pretty crucial. We'll break down the steps to calculate the amount of carbon dioxide (CO₂) produced when 3.2 grams of methane are burned with 16 grams of oxygen (O₂). Trust me, once you get the hang of this, you'll be able to solve all sorts of stoichiometry problems!

Understanding the Chemical Equation

The first step in any stoichiometry problem is understanding the chemical equation. Our equation for the combustion of methane looks like this:

CH₄(g) + O₂(g) → CO₂(g) + H₂O(g)

This equation tells us that methane gas (CH₄) reacts with oxygen gas (O₂) to produce carbon dioxide gas (CO₂) and water (H₂O). But here's the catch: this equation isn't balanced! A balanced chemical equation is absolutely essential because it follows the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. This means we need to have the same number of atoms of each element on both sides of the equation.

So, let's balance it! We can see that we have one carbon atom on each side, which is great. However, we have four hydrogen atoms on the left and only two on the right. To fix this, we can add a coefficient of 2 in front of the H₂O:

CH₄(g) + O₂(g) → CO₂(g) + 2H₂O(g)

Now, we have four hydrogen atoms on both sides. But look! We've changed the number of oxygen atoms. We now have two oxygen atoms in CO₂ and two in 2H₂O, totaling four oxygen atoms on the right. On the left, we only have two. To balance the oxygen, we add a coefficient of 2 in front of the O₂:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

Now, we have a balanced equation! We have one carbon atom, four hydrogen atoms, and four oxygen atoms on both sides. This balanced equation tells us the mole ratio in which the reactants react and the products are formed. In this case, one mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and two moles of water. This mole ratio is the key to solving our problem.

Converting Grams to Moles

The balanced equation deals with moles, not grams. So, the next crucial step is to convert the given masses of methane and oxygen into moles. To do this, we'll need the molar masses of CH₄ and O₂. Remember, molar mass is the mass of one mole of a substance and is numerically equal to its atomic or molecular weight in grams per mole (g/mol).

The problem gives us the atomic masses: Ar (C) = 12, O = 16, and H = 1. Let's calculate the molar masses:

• Molar mass of CH₄: 1 carbon (1 * 12 g/mol) + 4 hydrogen (4 * 1 g/mol) = 12 + 4 = 16 g/mol
• Molar mass of O₂: 2 oxygen (2 * 16 g/mol) = 32 g/mol

Now we can convert the given masses to moles using the formula:

Moles = Mass (g) / Molar mass (g/mol)

For methane (CH₄):

Moles of CH₄ = 3.2 g / 16 g/mol = 0.2 moles

For oxygen (O₂):

Moles of O₂ = 16 g / 32 g/mol = 0.5 moles

So, we have 0.2 moles of methane and 0.5 moles of oxygen. But which one will determine how much CO₂ we can actually produce? This leads us to the concept of the limiting reactant.

Identifying the Limiting Reactant

The limiting reactant is the reactant that is completely consumed in a chemical reaction. It's like the ingredient you run out of first when you're baking a cake. The limiting reactant determines the maximum amount of product that can be formed. The other reactant(s) are called the excess reactants, meaning there will be some leftover after the reaction is complete.

To find the limiting reactant, we need to compare the mole ratio of the reactants to the stoichiometric ratio from the balanced equation. Our balanced equation is:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

This tells us that 1 mole of CH₄ reacts with 2 moles of O₂. Let's see if our amounts match this ratio. We have 0.2 moles of CH₄. According to the stoichiometry, we would need 2 * 0.2 = 0.4 moles of O₂ to react completely with the methane. We have 0.5 moles of O₂, which is more than we need. This means that methane is the limiting reactant because it will run out first, and oxygen is the excess reactant.

Another way to think about it is to calculate how much CO₂ each reactant could produce if it were the limiting reactant.

• If CH₄ was the limiting reactant (0.2 moles), we could produce 0.2 moles of CO₂ (because the mole ratio between CH₄ and CO₂ is 1:1).
• If O₂ was the limiting reactant (0.5 moles), we could produce 0.5 moles O₂ / 2 = 0.25 moles of CO₂ (because the mole ratio between O₂ and CO₂ is 2:1).

The reactant that can produce the least amount of product is the limiting reactant. In this case, CH₄ can produce less CO₂ (0.2 moles) than O₂ (0.25 moles), confirming that CH₄ is the limiting reactant.

Calculating the Moles of CO₂ Formed

Now that we know methane is the limiting reactant, we can use its amount to calculate the amount of CO₂ formed. From the balanced equation, the mole ratio between CH₄ and CO₂ is 1:1. This means that for every 1 mole of CH₄ that reacts, 1 mole of CO₂ is produced.

Since we have 0.2 moles of CH₄, we will produce 0.2 moles of CO₂.

This step is straightforward once you've identified the limiting reactant. The key is to use the stoichiometric coefficients from the balanced equation to establish the mole ratio between the limiting reactant and the product you're interested in.

Converting Moles of CO₂ to Grams

The question asks for the mass of CO₂ in grams, so our final step is to convert moles of CO₂ to grams. To do this, we'll use the molar mass of CO₂. Let's calculate it:

• Molar mass of CO₂: 1 carbon (1 * 12 g/mol) + 2 oxygen (2 * 16 g/mol) = 12 + 32 = 44 g/mol

Now we can use the formula:

Mass (g) = Moles * Molar mass (g/mol)

Mass of CO₂ = 0.2 moles * 44 g/mol = 8.8 grams

Final Answer

So, the mass of CO₂ formed when 3.2 grams of methane are burned with 16 grams of oxygen is 8.8 grams.

Key Takeaways for Stoichiometry Success

Guys, that was a full stoichiometry problem! Let's recap the key steps to make sure you've got it:

1. Balance the chemical equation: This is the foundation of stoichiometry. Make sure you have the same number of atoms of each element on both sides of the equation.
2. Convert grams to moles: Use molar masses to convert the given masses of reactants into moles. Moles are the currency of chemical reactions.
3. Identify the limiting reactant: Determine which reactant will run out first. This reactant dictates the maximum amount of product that can be formed.
4. Calculate moles of product: Use the mole ratio from the balanced equation to find the moles of the desired product formed from the limiting reactant.
5. Convert moles of product to grams: Use the molar mass of the product to convert moles back into grams if the question asks for mass.

Stoichiometry might seem tricky at first, but with practice, you'll become a pro. Remember to break down the problem into smaller steps, and always double-check your work. Keep practicing, and you'll nail those chemistry exams! Good luck!