Question

Methane, \(\mathrm{CH}_{4},\) burns in oxygen. (a) What are the products of the reaction? (b) In 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?

Short 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) 101.7 g of \(\mathrm{O}_2\). (d) 127.2 g of products.

Step-by-step solution

Identify the Products

When methane (\(\mathrm{CH}_4\)) combusts in oxygen (\(\mathrm{O}_2\)), the products are carbon dioxide (\(\mathrm{CO}_2\)) and water (\(\mathrm{H}_2\mathrm{O}\)). This is a typical combustion reaction where hydrocarbons react with oxygen to produce carbon dioxide and water.

Write the Unbalanced Equation

The unbalanced chemical equation for the combustion of methane in oxygen is:\[ \mathrm{CH}_4 + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O} \]

Balance the Chemical Equation

To balance the chemical equation, count the atoms of each element on both sides:1. Methane has 1 carbon atom, so we need 1 \(\mathrm{CO}_2\) molecule.2. Methane has 4 hydrogen atoms, so we need 2 \(\mathrm{H}_2\mathrm{O}\) molecules.3. Oxygen appears in both products: 2 atoms in \(\mathrm{CO}_2\) and 2 atoms in \(\mathrm{H}_2\mathrm{O}\), totaling 4 oxygen atoms from 2 \(\mathrm{O}_2\) molecules.The balanced equation is:\[ \mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O} \]

Calculate the Molar Mass of Methane

The molar mass of methane (\(\mathrm{CH}_4\)) is calculated by adding the molar masses of its atoms:- Carbon: 12.01 g/mol- Hydrogen: 1.01 g/mol each for 4 atomsMolar mass of \(\mathrm{CH}_4 = 12.01 + (4 \times 1.01) = 16.05 \text{ g/mol}\).

Calculate Moles of Methane

Determine the moles of methane in \(25.5\, \mathrm{g}\) using its molar mass:\[ \text{Moles of } \mathrm{CH}_4 = \frac{25.5\, \mathrm{g}}{16.05\, \text{g/mol}} = 1.589 \text{ mol} \]

Calculate Moles of Oxygen Required

From the balanced equation, the combustion of 1 mole of \(\mathrm{CH}_4\) requires 2 moles of \(\mathrm{O}_2\). Thus, the moles of \(\mathrm{O}_2\) required for 1.589 moles of \(\mathrm{CH}_4\) is:\[ 1.589 \text{ mol} \times 2 = 3.178 \text{ mol of } \mathrm{O}_2 \]

Calculate the Mass of Oxygen Required

Calculate the mass of 3.178 moles of \(\mathrm{O}_2\) using the molar mass of \(\mathrm{O}_2\) (32.00 g/mol):\[ \text{Mass of } \mathrm{O}_2 = 3.178 \text{ mol} \times 32.00 \text{ g/mol} = 101.7 \text{ g} \]

Calculate Total Mass of Products

The combustion is a closed system where mass is conserved. Therefore, the total mass of the products equals the total mass of the reactants (methane + oxygen).Mass of \(\mathrm{CH}_4 + \) Mass of \(\mathrm{O}_2 = 25.5 \text{ g} + 101.7 \text{ g} = 127.2 \text{ g} \)

Key concepts

Balanced Chemical Equation

A balanced chemical equation is vital in representing a chemical reaction accurately. It ensures that the number of each type of atom is the same on the reactant side as on the product side. For the combustion of methane, we begin with an unbalanced equation: \[ \mathrm{CH}_4 + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O} \] To balance it, we need to consider each element:- **Carbon**: One carbon atom in methane requires one \( \mathrm{CO}_2 \)- **Hydrogen**: There are four hydrogen atoms in methane, so we need two \( \mathrm{H}_2\mathrm{O} \) molecules.- **Oxygen**: The products (\( \mathrm{CO}_2 \) and \( \mathrm{H}_2\mathrm{O} \)) contain four oxygen atoms in total, requiring two \( \mathrm{O}_2 \) molecules. Thus, the balanced equation is: \[ \mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O} \] Balancing equations is crucial because it reflects the law of conservation of mass, stating that matter cannot be created or destroyed in a chemical reaction.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance. It helps us convert between grams and moles, which is essential in stoichiometry. To find the molar mass of methane \((\mathrm{CH}_4)\), consider the atomic masses of carbon and hydrogen:- **Carbon (C)**: 12.01 g/mol- **Hydrogen (H)**: 1.01 g/molMethane has one carbon atom and four hydrogen atoms, so its molar mass is:\[ \text{Molar mass of } \mathrm{CH}_4 = 12.01 + 4 \times 1.01 = 16.05 \text{ g/mol} \]This calculation helps us understand how much one mole of methane weighs, which is crucial when performing further calculations involving the reaction.

Stoichiometry

Stoichiometry is an essential concept in chemistry used to compute the quantities of reactants and products in a chemical reaction. Using the Balanced Chemical Equation, we can determine the relationship between the amounts of different substances:For the combustion of methane: \[ \mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O} \]From this equation, we see that:- **1 mole of methane** reacts with **2 moles of oxygen** to produce 1 mole of carbon dioxide and 2 moles of water.To find out how much oxygen is needed to combust 25.5 g of methane, convert the mass of methane to moles:\[ \text{Moles of } \mathrm{CH}_4 = \frac{25.5\, \mathrm{g}}{16.05\, \text{g/mol}} = 1.589 \text{ mol} \]According to the balanced equation, **1.589 moles of methane** require twice the moles of oxygen:\[ 1.589 \times 2 = 3.178 \text{ moles of } \mathrm{O}_2 \]Using stoichiometry, you can understand and calculate the proportions of reactants and products in any chemical reaction accurately.

Chemical Reaction Products

When methane combusts in oxygen, it typically forms carbon dioxide \((\mathrm{CO}_2)\) and water \((\mathrm{H}_2\mathrm{O})\). Identifying the chemical reaction products is fundamental in understanding any reaction's outcome:- **Carbon Dioxide (CO\(_2\))** is a common product of combustion reactions involving hydrocarbons.- **Water (H\(_2\)O)** is another typical byproduct when hydrogen burns in the presence of oxygen.The total mass of the products from a chemical reaction is equal to the total mass of the reactants, as illustrated by the conservation of mass principle:In the combustion of methane, the total mass before and after the reaction remains the same:- **Total mass of reactants** is the sum of the mass of methane and oxygen: - Methane: 25.5 g - Oxygen: 101.7 g (calculated previously)- This leads to the **total mass of products** equaling 127.2 g.By understanding the products, you can apply these principles to anticipate the results of other similar combustion reactions.