Problem 5 - Chemistry and Chemical Reactivity

Question

Methane, \(\mathrm{CH}_{4},\) burns in oxygen. (a) What are the products of the reaction? (b) Write the balanced equation for the reaction. (c) What mass of \(\mathrm{O}_{2}\), in grams, is required for complete combustion of \(25.5 \mathrm{g}\) of methane? (d) What is the total mass of products expected from the combustion of \(25.5 \mathrm{g}\) of methane?

Step-by-Step Solution

Answer

(a) Carbon dioxide and water. (b) \(\mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}\). (c) 101.76 grams. (d) 127.26 grams.

Step 1: Identify Reaction Products

When methane (\(\mathrm{CH}_4\)) burns in oxygen (\(\mathrm{O}_2\)), it undergoes a complete combustion reaction. The products of this reaction are carbon dioxide (\(\mathrm{CO}_2\)) and water (\(\mathrm{H}_2\mathrm{O}\)).

Step 2: Write the Balanced Equation

The balanced chemical equation for the combustion of methane is:\[\mathrm{CH}_4(g) + 2\mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) + 2\mathrm{H}_2\mathrm{O}(l)\]This equation shows that one molecule of methane reacts with two molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water.

Step 3: Calculate Moles of Methane

To find out how much oxygen is needed, first calculate the number of moles of methane in 25.5 grams. The molar mass of methane is 16.04 g/mol.\[\text{Moles of } \mathrm{CH}_4 = \frac{25.5 \text{ g}}{16.04 \text{ g/mol}} \approx 1.59 \text{ moles}\]

Step 4: Determine Moles of Oxygen Required

Using the stoichiometry of the balanced equation, 1 mole of \(\mathrm{CH}_4\) requires 2 moles of \(\mathrm{O}_2\). Thus, 1.59 moles of \(\mathrm{CH}_4\) requires:\[\text{Moles of } \mathrm{O}_2 = 1.59 \text{ moles } \times 2 = 3.18 \text{ moles}\]

Step 5: Calculate Mass of Oxygen Required

Convert the moles of oxygen to grams. The molar mass of oxygen (\(\mathrm{O}_2\)) is 32.00 g/mol:\[\text{Mass of } \mathrm{O}_2 = 3.18 \text{ moles} \times 32.00 \text{ g/mol} = 101.76 \text{ grams}\]

Step 6: Calculate Total Mass of Products

Using the law of conservation of mass, the mass of the reactants will equal the mass of the products. Therefore, the total mass of products is the sum of the mass of methane and the mass of oxygen:\[\text{Total mass of products} = 25.5 \text{ g (methane)} + 101.76 \text{ g (oxygen)} = 127.26 \text{ grams}\]

Key Concepts

Chemical Equations

Chemical equations are the language of chemistry. They depict how substances react with each other to form new compounds. In the case of combustion reactions, like that of methane (\(\mathrm{CH}_4\)), the chemical equation will show reactants on the left side and products on the right. For the combustion of methane, the equation is:\[\mathrm{CH}_4(g) + 2\mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) + 2\mathrm{H}_2\mathrm{O}(l)\]

• The substances before the arrow are the reactants: methane and oxygen.
• The substances after the arrow are the products: carbon dioxide and water.

This equation tells us that one molecule of methane reacts with two molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water.
The reaction is a clear display of how reactants are transformed through a chemical change, indicating not just what happens, but how molecules rearrange their bonds to form new substances.

Stoichiometry

Stoichiometry is the mathematical part of chemistry that deals with the relative quantities of reactants and products in chemical reactions. In the balanced equation of methane combustion, we see that 1 mole of methane needs 2 moles of oxygen. This is a direct stoichiometric relationship, derived from balancing the atoms on both sides of the equation.
Here's how stoichiometry helps in calculations:

• It helps determine how much of each reactant is needed and how much of each product will be formed.
• For instance, if we have 1 mole of methane, stoichiometry tells us we need 2 moles of oxygen to completely react with it.
• Through stoichiometry, from 1 mole of methane, we produce 1 mole of carbon dioxide and 2 moles of water.

Understanding stoichiometry is crucial for determining the proportions of elements involved and predicting the outcomes of reactions.

Molar Mass

Molar mass is essential when converting between moles and grams, especially chemical reactions. It represents the mass of one mole of a substance, given in g/mol. For example, in the methane combustion reaction:

• Molar mass of methane (\(\mathrm{CH}_4\)) is 16.04 g/mol.
• Molar mass of oxygen (\(\mathrm{O}_2\)) is 32.00 g/mol.

To calculate how much of a substance is needed or produced in grams, you'll need to know its molar mass. Assume you need to find out how much oxygen is required to burn 25.5 g of methane.
First, calculate moles of methane:\[\text{Moles of } \mathrm{CH}_4 = \frac{25.5 \text{ g}}{16.04 \text{ g/mol}} \approx 1.59 \text{ moles}\]
This value is then used to determine how many moles of oxygen are needed using stoichiometry. Understanding molar mass is critical for practical applications in laboratory settings, ensuring precise measurement and chemical reaction compliance.

Balanced Equations

Balanced equations ensure the law of conservation of mass is followed, showing equal atoms for each element on both sides of the equation. In the methane combustion equation, balancing is achieved by adjusting coefficients:\[\mathrm{CH}_4(g) + 2\mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) + 2\mathrm{H}_2\mathrm{O}(l)\]

• The equation indicates:
  • 1 carbon atom on both sides.
  • 4 hydrogen atoms on both sides.
  • 4 oxygen atoms on both sides.

Each type of atom is conserved, fulfilling the law of conservation of mass.
Balanced equations allow chemists to understand the exact proportions of reactants required and products obtained. It's essential for efficient resource use and accuracy in experiments. Ensuring that equations are properly balanced also prevents errors in calculations surrounding chemical manufacturing and processes.