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: Calculate Moles of Methane

Given the molecular weight of methane (CH extsubscript{4}) is 16 g/mol, we can find the number of moles of CH extsubscript{4} using the formula \( \,=\frac{\text{mass given}}{\text{molar mass}} \,\). Thus, \( \,n=\frac{32}{16} = 2\,\text{moles of CH}_4\).

Step 2: Write Combustion Reaction

For the complete combustion of CH extsubscript{4}, the balanced chemical equation is:\[\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}\]This indicates that 1 mole of CH extsubscript{4} requires 2 moles of O extsubscript{2} for complete combustion.

Step 3: Calculate Moles of Required Oxygen

Since 1 mole of CH extsubscript{4} requires 2 moles of O extsubscript{2}, for 2 moles of CH extsubscript{4}, we need \(2 \times 2 = 4\) moles of O extsubscript{2}.

Step 4: Volume of Oxygen at STP

At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore, 4 moles of O extsubscript{2} will occupy \(4 \times 22.4 = 89.6\) liters.

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that allows us to predict the amounts of reactants and products involved in a chemical reaction. This concept is based on the idea that chemical reactions occur in fixed ratios as dictated by their balancing equations. When working with stoichiometry, it's important to first write the balanced chemical equation, which shows the relationship between the reactants and products.

Understanding stoichiometry entails

• Recognizing the mole ratio from the balanced equation.
• Determining how these ratios translate to quantities in grams, liters, or moles.
• Applying these ratios to find unknown values like necessary reactants or formed products.

In the exercise involving methane (\( ext{CH}_4\)), we use the balanced equation of\( ext{CH}_4 + 2 ext{O}_2 ightarrow ext{CO}_2 + 2 ext{H}_2 ext{O}\)to understand that one mole of methane burns fully with two moles of oxygen.

Moles Calculation

Moles provide a way to connect the microscopic world of atoms and molecules to measurable quantities in a laboratory setting. A mole is a unit of measurement that represents a specific number of particles, known as Avogadro's number, which is approximately\(6.022 imes 10^{23}\) particles.

Calculating moles requires using the formula:\[ ext{Moles} = \frac{\text{Mass of substance in grams}}{\text{Molar mass of the substance in g/mol}}.\]In the exercise, methane (\( ext{CH}_4\)) has a molar mass of 16 g/mol. If 32 grams of methane are given, the calculation becomes:\[ ext{Moles of } \text{CH}_4 = \frac{32}{16} = 2 ext{ moles}\]This tells us we are working with two moles of methane, and forms the basis for further stoichiometric calculations.

Standard Temperature and Pressure (STP)

Understanding Standard Temperature and Pressure (STP) is crucial when comparing the properties of gases. STP is defined as a temperature of 0°C (273.15 K) and pressure of 1 atm (101.325 kPa), and it provides a reference point to define the behavior of gases under these conditions.

This standard is vital because it simplifies calculations and the understanding of gas behaviors. Gases behave predictably under STP, and their volumes can be compared directly, since variations due to temperature and pressure are eliminated. Knowing if a problem states conditions at STP allows you to use specific, standardized values, such as the default volume one mole of gas occupies.

Gas Volume at STP

At Standard Temperature and Pressure (STP), any gas will occupy a volume of 22.4 liters per mole. This concept is rooted in Avogadro's law, which posits that equal volumes of gases, at the same temperature and pressure, contain an equal number of particles.

For the oxygen gas required in the combustion reaction of methane, knowing that oxygen gas at STP occupies 22.4 liters per mole simplifies our calculations greatly:

• Determine how many moles are required by using the stoichiometric ratios from the balanced equation.
  • In the exercise, 4 moles of \( ext{O}_2\) are needed.
• Calculate the volume by multiplying the number of moles by 22.4 L/mole.
  • Thus, \(4 imes 22.4 = 89.6 ext{ liters of } ext{O}_2\).

This provides the volume of oxygen needed and highlights the important role STP conditions play in gas volume calculations.