Question

A \(5.72-\mathrm{g}\) sample of graphite was heated with \(68.4 \mathrm{~g}\) of \(\mathrm{O}_{2}\) in a \(8.00-\mathrm{L}\) flask. The reaction that took place was $$ \mathrm{C} \text { (graphite) }+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) $$ After the reaction was complete, the temperature in the flask was \(182^{\circ} \mathrm{C}\). What was the total pressure inside the flask?

Short answer

Total pressure is 2.23 atm inside the flask.

Step-by-step solution

Identify Given Values

We are given a 5.72 g sample of graphite and 68.4 g of O₂. The volume of the flask is 8.00 L and the temperature after the reaction is complete is 182°C.

Convert Temperature to Kelvin

To use the ideal gas law, convert the temperature from Celsius to Kelvin: \[ T(K) = 182 + 273.15 = 455.15 \, \text{K} \]

Determine Moles of Reactants

Calculate moles of graphite (C): Since the molar mass of carbon (graphite) is about 12.01 g/mol, \[ n_C = \frac{5.72}{12.01} \approx 0.476 \text{ mol} \]Calculate moles of O₂:Molar mass of O₂ is 32.00 g/mol,\[ n_{O_2} = \frac{68.4}{32.00} \approx 2.14 \text{ mol} \]

Identify Limiting Reactant

The reaction consuming the stoichiometry 1:1 ratio of C to O₂ forms CO₂. Since we have 0.476 mol of C and 2.14 mol of O₂, graphite is the limiting reactant.

Calculate Moles of CO₂ Formed

Since graphite is the limiting reactant and reacts 1:1 with O₂ to form CO₂, the moles of CO₂ formed is equal to moles of C, which is 0.476 mol.

Calculate Total Moles of Gas After Reaction

After the reaction, the remaining gas is CO₂ (0.476 mol), with no unreacted O₂ left as we calculated moles based on limiting reagent. So, total moles: 0.476 mol.

Apply Ideal Gas Law to Find Pressure

Use the Ideal Gas Law, \[ PV = nRT \]where \( R = 0.0821 \, \text{L atm/mol K} \). Solve for \( P \):\[ P = \frac{nRT}{V} = \frac{0.476 \times 0.0821 \times 455.15}{8.00} \approx 2.23 \, \text{atm} \]

Key concepts

Understanding Stoichiometry

Stoichiometry is like a recipe for chemical reactions. It tells us the proportions of reactants and products involved in a reaction. In the given exercise, the reaction between graphite (C) and oxygen (\(\mathrm{O}_2\)) to form carbon dioxide (\(\mathrm{CO}_2\)) is described by the equation: \[\mathrm{C} + \mathrm{O}_2 \rightarrow \mathrm{CO}_2\]

• This equation indicates a 1:1 mole ratio between graphite and oxygen, meaning for each mole of graphite, one mole of oxygen is needed to produce one mole of carbon dioxide.
• Knowing the stoichiometric coefficients is essential for understanding how much of each substance is involved.

It's crucial to understand stoichiometry to determine how much of each reactant is needed or how much product we can expect.

Finding the Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that gets completely consumed first, thus determining the maximum amount of product that can be formed.

• From our example: Graphite and oxygen react in a 1:1 ratio, meaning ideally, we need equal moles of each to complete the reaction without leftovers.
• With 0.476 moles of graphite and 2.14 moles of oxygen, graphite will run out first and limit the production of carbon dioxide.
• Identifying the limiting reactant helps in maximizing product yield and is crucial for efficient resource utilization.

Calculating Moles

The concept of moles is fundamental in chemistry for expressing amounts of a chemical substance.

• The mole is based on Avogadro's number, \(6.022 \times 10^{23}\) entities of the substance.
• In our exercise:
  • Graphite is calculated as \( \frac{5.72 \, g}{12.01 \, g/mol} \approx 0.476 \, \text{moles} \).
  • Oxygen is calculated as \( \frac{68.4 \, g}{32.00 \, g/mol} \approx 2.14 \, \text{moles} \).
• Converting mass to moles allows us to use stoichiometry to predict reaction outcomes.

Temperature Conversion: Celsius to Kelvin

Temperature conversion is essential when working with gases and the ideal gas law. The Kelvin scale is used because it provides an absolute reference point where zero represents a complete absence of thermal energy.

• The relationship between Celsius and Kelvin is straightforward: add 273.15 to the Celsius temperature.
• In this case, 182°C becomes 455.15 K after conversion.
• This conversion is vital for accurately using the ideal gas law: \(PV = nRT\), where temperature must be in Kelvin units.

Accurate temperature conversion ensures that calculations of gas behavior, like pressure and volume, are correct.