Problem 85 - Chemistry Matter and Change

Question

A 6.5 -L sample of hydrogen sulfide is treated with a catalyst to promote the reaction shown below. $$2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{S}(\mathrm{s})$$ If the \(\mathrm{H}_{2} \mathrm{S}\) reacts completely at 2.0 atm and \(290 \mathrm{K},\) how much water vapor, in grams, is produced?

Step-by-Step Solution

Answer

Approximately 9.55 grams of water vapor ( \(\mathrm{H}_2 \mathrm{O}\n\)) is produced.

Step 1: Understand the Reaction and Conditions

The chemical reaction involves hydrogen sulfide (\(\mathrm{H}_2 \mathrm{S}\)) reacting completely with oxygen gas to form water vapor and solid sulfur. The volume of \(\mathrm{H}_2 \mathrm{S} \) given is 6.5 L, with initial conditions of 2.0 atm pressure and 290 K temperature.

Step 2: Use the Ideal Gas Law to find Moles of H2S

The ideal gas law is \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is moles of gas, \(R\) is the ideal gas constant (0.0821 L atm K\(^{-1}\) mol\(^{-1}\)), and \(T\) is temperature in Kelvin. Substitute the known values: \( (2.0\, \mathrm{atm}) \times (6.5\, \mathrm{L}) = n \times (0.0821\, \mathrm{L} \cdot \mathrm{atm} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}) \times (290\, \mathrm{K}) \). Solve for \(n\).

Step 3: Calculate the Moles of Water Produced

From the balanced chemical equation, each mole of \(\mathrm{H}_2 \mathrm{S}\) produces one mole of \(\mathrm{H}_2 \mathrm{O}\). Therefore, the moles of water generated will be equal to the moles of \(\mathrm{H}_2 \mathrm{S}\) that reacted completely, which we calculated in Step 2.

Step 4: Convert Moles of Water to Grams

The molecular weight of water (\(\mathrm{H}_2 \mathrm{O}\)) is approximately 18.015 g/mol. Multiply the moles of water calculated in Step 3 by the molecular weight of water to find the mass in grams. Formula: mass = moles \(\times\) molecular weight.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a key equation that helps us understand how gases behave under different conditions. It's a cornerstone in chemistry and involves the relationship between pressure ( \( P \)), volume ( \( V \)), temperature ( \( T \)), and the amount of gas in moles ( \( n \)). The Ideal Gas Law is represented as:
\[PV = nRT\]Here, \( R \) is the ideal gas constant with a value of 0.0821 L atm K\(^{-1}\) mol\(^{-1}\).
This equation allows us to calculate the number of moles ( \( n \)) of a gas when we know its pressure, volume, and temperature. For example, if you have a gas at a given pressure and volume, and you know the temperature, you can rearrange the equation to solve for \( n \):
\[n = \frac{PV}{RT}\]This is how in our problem, we determined the moles of hydrogen sulfide (\( \mathrm{H}_2 \mathrm{S} \)) present in the container.

Chemical Reactions

Chemical reactions describe how substances interact to form new products. In the exercise, hydrogen sulfide (\( \mathrm{H}_2 \mathrm{S} \)) reacts with oxygen (\( \mathrm{O}_2 \)) to form water vapor and sulfur. This is a typical combustion reaction where reactants bond to release new compounds and energy.
The reaction equation is balanced, meaning the number of atoms for each element is equal on both sides. For example:

• 2 molecules of \( \mathrm{H}_2 \mathrm{S} \) react with 1 molecule of \( \mathrm{O}_2 \)
• Producing 2 molecules of \( \mathrm{H}_2 \mathrm{O} \) and 2 atoms of sulfur

In chemical reactions, balancing is essential because it conserves mass and complies with the law of conservation of matter. This understanding helps predict the amounts of products formed, as seen in our reaction where each mole of hydrogen sulfide leads to a mole of water.

Mole Calculations

Moles are a fundamental concept in chemistry that bridge the gap between grams and number of atoms or molecules. In our exercise, we used moles to connect the volume and pressure of a gas to the mass of products formed.
For the hydrogen sulfide, we used the Ideal Gas Law to calculate its moles. From the chemical reaction, we know that the moles of hydrogen sulfide directly translate to the moles of water produced:

• 1 mole of \( \mathrm{H}_2 \mathrm{S} \) produces 1 mole of \( \mathrm{H}_2 \mathrm{O} \)

Once we have the moles of water, we convert it to grams using its molecular weight (around 18.015 g/mol). This is done by multiplying the number of moles by the molecular weight:\[\text{mass} = \text{moles} \times \text{molecular weight}\]Using this approach, we are able to find out exactly how much water vapor, in grams, results from the reaction.

Gas Laws

Gas laws are an essential component of chemistry that explore the behavior of gases. They include a series of physical laws that describe the interrelations between pressure, volume, and temperature of a gas. The Ideal Gas Law often summarizes these relationships but understanding each concept separately gives a better glimpse into gas behavior.
For instance:

• Boyle's Law: States that pressure is inversely proportional to volume at constant temperature (\( PV = \text{constant} \)).
• Charles's Law: Indicates that volume is directly proportional to temperature at constant pressure (\( V/T = \text{constant} \)).
• Avogadro's Law: Asserts that volume is proportional to the number of moles at constant temperature and pressure (\( V/n = \text{constant} \)).

These laws set the groundwork for predicting how a gas will react to changes in its environment. In our exercise, they help clarify why changing one parameter, like pressure or temperature, alters the amount of product gas formed.