Calculating CO₂ Production from Heating Carbon

This question asks us to determine the amount of carbon dioxide (\(\text{CO}_2\)) produced when a specific amount of carbon (\(\text{C}\)) is heated. This involves a basic chemical reaction known as combustion, where carbon reacts with oxygen to form carbon dioxide. We can use stoichiometry, the study of the quantitative relationships or ratios of substances involved in chemical reactions, to solve this problem.

Chemical Equation for Carbon Combustion

The balanced chemical equation for the complete combustion of carbon is:

\(\text{C} + \text{O}_2 \rightarrow \text{CO}_2\)

This equation shows that one mole of carbon reacts with one mole of oxygen gas to produce one mole of carbon dioxide.

Determining Molar Masses

To work with masses in stoichiometry, we need the molar masses of the substances involved. The molar mass of an element or compound is the mass of one mole of that substance.

• Molar mass of Carbon (\(\text{C}\)) = \(12.01 \, \text{g/mol}\) (approximately \(12 \, \text{g/mol}\))
• Molar mass of Oxygen (\(\text{O}_2\)) = \(2 \times \text{Atomic mass of O} = 2 \times 16.00 \, \text{g/mol} = 32.00 \, \text{g/mol}\)
• Molar mass of Carbon Dioxide (\(\text{CO}_2\)) = \(\text{Molar mass of C} + 2 \times \text{Atomic mass of O} = 12.01 \, \text{g/mol} + 2 \times 16.00 \, \text{g/mol} = 12.01 + 32.00 \, \text{g/mol} = 44.01 \, \text{g/mol}\) (approximately \(44 \, \text{g/mol}\))

For simplicity in calculations, we often use rounded atomic masses. Let's use \(12 \, \text{g/mol}\) for C and \(44 \, \text{g/mol}\) for \(\text{CO}_2\).

Stoichiometric Calculation

From the balanced equation, \(1\) mole of \(\text{C}\) produces \(1\) mole of \(\text{CO}_2\). Using the molar masses, this means \(12 \, \text{g}\) of \(\text{C}\) produces \(44 \, \text{g}\) of \(\text{CO}_2\).

We are given \(1 \, \text{kg}\) of carbon, which is equal to \(1000 \, \text{g}\). We can set up a ratio based on the stoichiometry:

\(\frac{\text{Mass of C}}{\text{Mass of CO}_2} = \frac{\text{Molar mass of C}}{\text{Molar mass of CO}_2}\)

Let \(x\) be the mass of \(\text{CO}_2\) produced from \(1000 \, \text{g}\) of \(\text{C}\).

\(\frac{1000 \, \text{g C}}{x \, \text{g CO}_2} = \frac{12 \, \text{g C}}{44 \, \text{g CO}_2}\)

Now, we solve for \(x\):

\(x = \frac{1000 \, \text{g C} \times 44 \, \text{g CO}_2}{12 \, \text{g C}}\)

\(x = \frac{1000 \times 44}{12} \, \text{g}\)

\(x = \frac{1000 \times 11}{3} \, \text{g}\) (by dividing 44 and 12 by 4)

\(x = \frac{11000}{3} \, \text{g}\)

The question asks for the mass in kilograms. Since \(1 \, \text{kg} = 1000 \, \text{g}\), we convert grams to kilograms by dividing by \(1000\):

Mass of \(\text{CO}_2\) in kg \(= \frac{11000/3 \, \text{g}}{1000 \, \text{g/kg}}\)

Mass of \(\text{CO}_2\) in kg \(= \frac{11000}{3 \times 1000} \, \text{kg}\)

Mass of \(\text{CO}_2\) in kg \(= \frac{11}{3} \, \text{kg}\)

Summary of Calculation Steps

1. Write and balance the chemical equation: \(\text{C} + \text{O}_2 \rightarrow \text{CO}_2\).
2. Determine molar masses: \(\text{C} \approx 12 \, \text{g/mol}\), \(\text{CO}_2 \approx 44 \, \text{g/mol}\).
3. Use stoichiometry: \(12 \, \text{g}\) C yields \(44 \, \text{g}\) \(\text{CO}_2\).
4. Convert given mass to grams: \(1 \, \text{kg} = 1000 \, \text{g}\).
5. Calculate \(\text{CO}_2\) mass in grams: \((\frac{44}{12}) \times 1000 \, \text{g} = \frac{11000}{3} \, \text{g}\).
6. Convert \(\text{CO}_2\) mass to kilograms: \(\frac{11000/3}{1000} \, \text{kg} = \frac{11}{3} \, \text{kg}\).

Therefore, heating \(1 \, \text{kg}\) of carbon produces \(\frac{11}{3} \, \text{kg}\) of carbon dioxide.

Revision Table: Key Concepts

Additional Information: Combustion and Mass Conservation

The reaction of carbon with oxygen is a common example of combustion. Combustion reactions typically produce heat and light. In this case, with sufficient oxygen, carbon undergoes complete combustion to form \(\text{CO}_2\). If there is insufficient oxygen, incomplete combustion can occur, producing carbon monoxide (\(\text{CO}\)) instead or along with \(\text{CO}_2\). The problem implies complete combustion.

According to the Law of Conservation of Mass, the total mass of the reactants in a closed system must equal the total mass of the products. In this reaction, the mass of carbon plus the mass of oxygen consumed will equal the mass of carbon dioxide produced. We calculated that \(1 \, \text{kg}\) (or \(1000 \, \text{g}\)) of carbon produces \(\frac{11000}{3} \, \text{g}\) of \(\text{CO}_2\). The mass of oxygen consumed would be the difference: \(\frac{11000}{3} \, \text{g} - 1000 \, \text{g} = \frac{11000 - 3000}{3} \, \text{g} = \frac{8000}{3} \, \text{g}\). Let's verify this with moles: \(1000 \, \text{g}\) C is \(\frac{1000}{12}\) moles. This reacts with \(\frac{1000}{12}\) moles of O2. Mass of O2 = \(\frac{1000}{12} \times 32 \, \text{g} = \frac{1000 \times 8}{3} \, \text{g} = \frac{8000}{3} \, \text{g}\). Mass of \(\text{CO}_2\) = \(\frac{1000}{12} \times 44 \, \text{g} = \frac{1000 \times 11}{3} \, \text{g} = \frac{11000}{3} \, \text{g}\). Indeed, \(1000 + \frac{8000}{3} = \frac{3000+8000}{3} = \frac{11000}{3}\), demonstrating mass conservation.