Problem 77 - Chemistry: Principles and Reactions

Question

A flask with a volume of \(10.0\) L contains \(0.400 \mathrm{~g}\) of hydrogen gas and \(3.20 \mathrm{~g}\) of oxygen gas. The mixture is ignited and the reaction $$ 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} $$ goes to completion. The mixture is cooled to \(27^{\circ} \mathrm{C}\). Assuming \(100 \%\) yield, (a) What physical state(s) of water is (are) present in the flask? (b) What is the final pressure in the flask? (c) What is the pressure in the flask if \(3.2 \mathrm{~g}\) of each gas is used?

Step-by-Step Solution

Answer

Answer: When 0.400 g of hydrogen gas and 3.20 g of oxygen gas react, the final pressure in the flask is approximately 0.734 atm, and the physical state of water is gaseous. When 3.2 g of each gas is used, the pressure in the flask is approximately 7.73 atm, and the physical state of water remains gaseous.

Step 1: Find the limiting reactant

Convert grams of hydrogen gas and oxygen gas to moles: - Using molar mass: H2 = 2.02 g/mol, O2 = 32 g/mol Moles of H2 = (0.400 g) / (2.02 g/mol) = 0.198 moles Moles of O2 = (3.20 g) / (32 g/mol) = 0.100 moles Now, compare the mole ratios: H2 reaction ratio = 0.198 / 2 (from balanced equation) O2 reaction ratio = 0.100 / 1 (from balanced equation) Since O2 has a smaller ratio, it is the limiting reactant.

Step 2: Calculate the theoretical yield of water

Using the balanced equation, 1 mole of O2 reacts with 2 moles of H2 to produce 2 moles of H2O. So, we can calculate the moles of H2O produced: Moles of H2O = 2 * moles of O2 = 2 * 0.100 = 0.200 moles

Step 3: Determine the remaining amount of reactant

Since H2 is not completely consumed in the reaction, calculate the remaining moles: Moles of H2 remaining = moles of H2 - 2 * moles of O2 = 0.198 - 2 * 0.100 = 0.098 moles

Step 4: Calculate the final pressure in the flask

Use the ideal gas law (PV = nRT) to find the pressure of the system. The total moles in the flask after the reaction are 0.098 moles of H2 remaining + 0.200 moles of H2O. Total moles = 0.098 + 0.200 = 0.298 moles Given volume of the flask is 10 L and after cooling the mixture temperature is 27°C = 27 + 273.15 = 300.15 K. R (ideal gas constant) = 0.0821 L atm / (K mol) Now, use PV = nRT to find the pressure (P): P = (nRT) / V = (0.298 mol * 0.0821 L atm / (K mol) * 300.15 K) / 10 L = 0.734 atm So, the final pressure in the flask is approximately 0.734 atm.

Step 5: Determine the physical state(s) of water

At the given temperature (27°C), water is in the gaseous state. Thus, only gaseous water is present in the flask.

Step 6: Calculate the pressure in the flask if 3.2 g of each gas is used

In this case, moles of H2 = (3.2 g) / (2.02 g/mol) = 1.58 moles Moles of O2 = (3.20 g) / (32 g/mol) = 0.100 moles Here, H2 is the limiting reactant. The theoretical yield of H2O: Moles of H2O = 2 * moles of H2 = 2 * 1.58 = 3.16 moles No O2 remains in the flask, and there are 3.16 moles of H2O present. Using the ideal gas law (PV = nRT), we can find the pressure: P = (nRT) / V = (3.16 mol * 0.0821 L atm / (K mol) * 300.15 K) / 10 L = 7.73 atm So, the pressure in the flask if 3.2 g of each gas is used is approximately 7.73 atm.

Key Concepts

ideal gas law

The ideal gas law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and amount of gas in a system. The equation is represented as \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles of gas, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. This law helps us understand how gases behave under different conditions.

To solve for any one property, we rearrange the equation based on the variables we know. For example, if the task is to find the pressure, we rearrange it to \( P = \frac{nRT}{V} \).

Here are some tips for using the ideal gas law effectively:

• Always convert temperature to Kelvin by adding 273.15 to the Celsius measurement.
• Make sure all units are consistent, typically using liters for volume and atmospheres for pressure.
• Apply the correct value for \( R \), which is 0.0821 L atm / (K mol) when using these units.

This law is particularly useful for finding unknowns after chemical reactions when the conditions of the gases change.

mole ratio

Mole ratio is a concept derived from a balanced chemical equation and represents the ratio of moles of one substance to the moles of another. It's a key tool in stoichiometry for predicting how much of a reactant might be needed or how much product will be formed.

For the reaction \( 2 \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \text{H}_2\text{O} \), the mole ratio is evident as:

• 2 moles of \( \text{H}_2 \) react with 1 mole of \( \text{O}_2 \).
• This produces 2 moles of \( \text{H}_2\text{O} \).

This implies that for every mole of oxygen, you can theoretically react it with two moles of hydrogen to produce water. If there are fewer moles of hydrogen or oxygen present, the one with lesser moles, according to the required mole ratio, becomes the limiting reactant.

Remember, determining the limiting reactant helps to ascertain the reactant that will be consumed first and thus, limits the extent of the reaction.

theoretical yield

Theoretical yield refers to the maximum amount of product that could be formed from a given amount of reactants, according to a chemical equation's ratio. It's calculated based on the limiting reactant principle and assumes that every single atom reacts perfectly.

To calculate the theoretical yield:

• First, determine the limiting reactant using mole ratios, as outlined previously.
• Convert the limiting reactant's quantity (in moles) to the amount of product based on the balanced equation's mole ratio.

In our example, with the balanced reaction \( 2 \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \text{H}_2\text{O} \), if \( 0.1 \) moles of \( \text{O}_2 \) acts as the limiting reactant, the theoretical yield of water is \( 0.2 \) moles.

This approach provides a benchmark for chemists to assess the efficiency of reactions by comparing the actual yield (what you actually get from an experiment) with the theoretical yield.

physical states of matter

Matter exists in different physical states – solid, liquid, gas, and sometimes plasma. These states have unique properties that determine how substances behave under various conditions.

In the context of the exercise, we're dealing with gases and the formation of water. The reaction yields water, and determining its state at room temperature (27°C) is crucial. Generally:

• Below 0°C, water solidifies into ice.
• Between 0°C and 100°C, it is typically in the liquid state.
• Above 100°C, water vapor or steam is present.

However, inside a closed flask, conditions like pressure can affect the state. Even at 27°C, water may remain in the vapor phase due to other factors such as pressure or initial gas conditions.

Understanding these states helps predict how chemicals and reactions behave under certain environmental conditions, which is vital for experimental planning and safety.