Problem 96 - Chemistry and Chemical Reactivity

Question

Methane is burned in a laboratory Bunsen burner to give \(\mathrm{CO}_{2}\) and water vapor. Methane gas is supplied to the burner at the rate of \(5.0 \mathrm{L} / \mathrm{min}\) (at a temperature of \(28^{\circ} \mathrm{C}\) and a pressure of \(773 \mathrm{mm} \mathrm{Hg}\) ). At what rate must oxygen be supplied to the burner (at a pressure of \(742 \mathrm{mm} \mathrm{Hg}\) and a temperature of \(\left.26^{\circ} \mathrm{C}\right) ?\)

Step-by-Step Solution

Answer

The oxygen must be supplied at a rate of 10.4 L/min.

Step 1: Write the Balanced Chemical Equation

The combustion of methane (\( \mathrm{CH}_4 \)) with oxygen (\( \mathrm{O}_2 \)) produces carbon dioxide (\( \mathrm{CO}_2 \)) and water (\( \mathrm{H}_2\mathrm{O} \)). The balanced equation is: \[\mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}\] This equation shows that 1 mole of methane reacts with 2 moles of oxygen.

Step 2: Apply the Ideal Gas Law to Methane

Use the ideal gas law \( PV = nRT \) to find the number of moles of methane. First, convert pressure from mmHg to atm: \( 773 \, \mathrm{mmHg} \times \frac{1 \, \mathrm{atm}}{760 \, \mathrm{mmHg}} = 1.017 \, \mathrm{atm} \). Convert the temperature to Kelvin: \( 28^{\circ} \mathrm{C} = 301 \mathrm{K} \).Using \( R = 0.0821 \, \mathrm{L} \, \mathrm{atm} / \mathrm{mol} \cdot \mathrm{K} \), substitute into the formula:\[ PV = nRT \quad \Rightarrow \quad (1.017 \, \mathrm{atm})(5.0 \, \mathrm{L}) = n (0.0821)(301 \, \mathrm{K})\]Solve for \( n \), the number of moles of methane:\[ n \approx 0.206 \, \mathrm{mol/min} \]

Step 3: Calculate Moles of Oxygen Needed

According to the balanced equation, 1 mole of methane requires 2 moles of oxygen. Therefore, the moles of oxygen needed per minute is:\[ 2 \times 0.206 \, \mathrm{mol/min} = 0.412 \, \mathrm{mol/min} \]

Step 4: Convert Temperature and Pressure for Oxygen

Convert pressure from mmHg to atm for oxygen: \( 742 \, \mathrm{mmHg} \times \frac{1 \, \mathrm{atm}}{760 \, \mathrm{mmHg}} = 0.976 \, \mathrm{atm} \). Convert the temperature to Kelvin \( 26^{\circ} \mathrm{C} = 299 \, \mathrm{K} \).

Step 5: Apply the Ideal Gas Law to Oxygen

Use the ideal gas law \( PV = nRT \) to find the volume of oxygen needed. Solve for \( V \):\[ V = \frac{nRT}{P} = \frac{(0.412 \, \mathrm{mol/min})(0.0821 \, \mathrm{L} \, \mathrm{atm/mol} \cdot \mathrm{K})(299 \, \mathrm{K})}{0.976 \, \mathrm{atm}} \]Calculate \( V \): \[ V \approx 10.4 \, \mathrm{L/min} \]

Step 6: Conclusion

From the calculations, oxygen should be supplied at a rate of approximately \( 10.4 \, \mathrm{L/min} \).

Key Concepts

Ideal Gas Law

The ideal gas law is a fundamental concept in chemistry that provides a relationship between pressure (\(P\)), volume (\(V\)), and temperature (\(T\)) of a gas with its number of moles (\(n\)). This relationship is expressed with the equation: \[PV = nRT\]where \(R\) is the universal gas constant, typically valued at \(0.0821 \, ext{L} \, ext{atm}/ ext{mol} \, ext{K}\).In the context of this problem, the ideal gas law helps us calculate the number of moles of gases involved in a chemical reaction. For methane, we used the given parameters (pressure in atm, volume in liters, and temperature in Kelvin) to determine the moles per minute. This data is crucial for further stoichiometric calculations, as it links the physical conditions to the chemical quantities required for reactions. Knowing how to apply the ideal gas law allows you to predict how changes in any of these variables affect others, which is vital when working with gases. It provides the basis for many calculations in fields like atmospheric science, engineering, and industrial processes.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products and are represented through balanced chemical equations. The equation for the combustion of methane is:\[\mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}\]This 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.Balancing chemical equations is essential to reflect the conservation of mass, ensuring that the number of atoms for each element is the same on both sides of the equation. This balance allows us to perform stoichiometric calculations to determine how much reactant is needed to produce a certain amount of product.Understanding this process gives chemists the ability to predict the outcome of reactions and scale them for practical uses, such as fuel combustion or synthesis of materials. It also helps in calculating yields and understanding the efficiency of reactions in various conditions.

Gas Laws

Gas laws are a set of principles that describe the behavior of gases under various conditions. These laws include Boyle's Law, Charles's Law, Avogadro's Law, and the Ideal Gas Law, which is a combination of these individual laws. In practical terms, gas laws allow us to predict how a gas will expand or compress when subjected to changes in temperature, pressure, or volume. For instance:

• **Boyle’s Law:** When temperature is constant, the pressure of a gas is inversely proportional to its volume.
• **Charles’s Law:** At constant pressure, the volume of a gas is directly proportional to its temperature in Kelvin.
• **Avogadro’s Law:** Equal volumes of gases, at the same temperature and pressure, have the same number of molecules.

Understanding these laws helps us manipulate and control gases in practical applications, such as hot air balloons or pressurized gas systems. In the given problem, gas laws provide the framework to calculate the oxygen supply needed to ensure complete combustion of methane, highlighting their real-world importance.