Problem 98 - A Complete Resource Book in Chemistry for JEE Main

Question

What is the volume (in litres) of oxygen at STP required for complete combustion of \(32 \mathrm{~g}\) of \(\mathrm{CH}_{4}\) ? (mol. wt of \(\mathrm{CH}_{4}=16\) ) (a) \(89.6\) (b) \(189.6\) (c) \(98.4\) (d) \(169.5\)

Step-by-Step Solution

Answer

The volume of oxygen required is 89.6 liters.

Step 1: Determine Moles of Methane

First, calculate the number of moles of methane, \(\mathrm{CH}_4\), given that the molecular weight of \(\mathrm{CH}_4\) is \(16 \text{ g/mol}\). Use the formula \[\text{Moles of } \mathrm{CH}_4 = \frac{\text{Mass of } \mathrm{CH}_4}{\text{Molar mass of } \mathrm{CH}_4}\]. For \(\text{Mass of } \mathrm{CH}_4 = 32 \text{ g}\), we get: \[ \text{Moles of } \mathrm{CH}_4 = \frac{32}{16} = 2 \, \text{moles}.\]

Step 2: Write Balanced Chemical Equation

The balanced chemical equation for the complete combustion of methane \(\mathrm{CH}_4\) is: \[ \mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}.\] It shows that 1 mole of \(\mathrm{CH}_4\) requires 2 moles of \(\mathrm{O}_2\) for complete combustion.

Step 3: Calculate Moles of Oxygen Required

According to the stoichiometric relationship provided by the balanced chemical equation, 2 moles of \(\mathrm{CH}_4\) will require \(2 \times 2 = 4\) moles of \(\mathrm{O}_2\) for complete combustion.

Step 4: Convert Moles of Oxygen to Volume

At standard temperature and pressure (STP), 1 mole of any ideal gas occupies \(22.4 \text{ L}\). Multiply the moles of \(\mathrm{O}_2\) by \(22.4 \text{ L/mol}\) to find the volume: \[ \text{Volume of } \mathrm{O}_2 = 4 \times 22.4 = 89.6 \text{ L}.\]

Key Concepts

Molar Mass Calculation

When performing chemical calculations, it is crucial to understand the concept of molar mass. Molar mass is the mass of one mole of a substance. It is usually expressed in grams per mole (g/mol). To find the molar mass of a compound, you add up the atomic masses of all the atoms in its molecular formula. For example, in our exercise dealing with methane (\(\mathrm{CH}_4\)), the molar mass is calculated by adding the atomic mass of one carbon atom and four hydrogen atoms:

• Carbon (C): about 12 g/mol
• Hydrogen (H) for four atoms: about 4 g/mol (1 g/mol each)

So, the molar mass of \(\mathrm{CH}_4\) is \(16 \, \text{g/mol}\). Molar mass is a vital step in converting the mass of a substance to the number of moles, which is a key step in solving many chemical problems.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantities of substances that participate in chemical reactions. It is based on understanding the balanced chemical equation, which shows the proportion of reactants and products. In our exercise, the combustion reaction for methane is:\[ \mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O} \]This equation tells us that one mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and two moles of water. Stoichiometry allows us to calculate the amount of reactants needed or products formed in a reaction. By knowing the moles of one reactant (methane in our case), we can use the molar ratios from the balanced equation to find the moles of another substance involved, such as oxygen, required for complete combustion.

Chemical Equation Balancing

Chemical equations must be balanced to conform with the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction. Balancing a chemical equation means ensuring that there are equal numbers of each type of atom on both sides of the equation. This is done by adjusting the coefficients, which are the numbers before the chemical formulas. For instance, in the complete combustion of methane:\[ \mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O} \]We see:

• One carbon atom on each side
• Four hydrogen atoms on each side
• Four oxygen atoms on each side

By ensuring the equation is balanced, we can accurately use it for stoichiometric calculations. Balancing equations is a fundamental skill for predicting the quantities of reactants or products involved in a chemical reaction.