Problem 48 - Chemistry and Chemical Reactivity

Question

If \(10.0 \mathrm{g}\) of carbon is combined with an exact, stoichiometric amount of oxygen \((26.6 \mathrm{g})\) to produce carbon dioxide, what is the theoretical yield of \(\mathrm{CO}_{2}\), in grams?

Step-by-Step Solution

Answer

The theoretical yield of \(\text{CO}_2\) is approximately 36.59 grams.

Step 1: Write the balanced chemical reaction

The reaction between carbon and oxygen to form carbon dioxide is represented by the balanced chemical equation: \[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \]

Step 2: Calculate the molar mass of reactants and products

First, calculate the molar masses of each of the reactants. The molar mass of carbon (C) is \(12.01 \, \text{g/mol}\). The molar mass of oxygen gas (\(\text{O}_2\)) is \(2 \times 16.00 = 32.00 \, \text{g/mol}\). The molar mass of carbon dioxide (\(\text{CO}_2\)) is \(12.01 + 2 \times 16.00 = 44.01 \, \text{g/mol}\).

Step 3: Determine moles of carbon and oxygen

Calculate the number of moles of carbon: \[ \text{moles of C} = \frac{10.0 \, \text{g}}{12.01 \, \text{g/mol}} \approx 0.833 \, \text{mol} \] Similarly, calculate the number of moles of oxygen: \[ \text{moles of } \text{O}_2 = \frac{26.6 \, \text{g}}{32.00 \, \text{g/mol}} \approx 0.831 \, \text{mol} \]

Step 4: Determine the limiting reactant

From the balanced chemical equation, it's clear that 1 mole of C reacts with 1 mole of \(\text{O}_2\) to produce 1 mole of \(\text{CO}_2\). Since the available moles of C are slightly higher than those of \(\text{O}_2\), \(\text{O}_2\) is the limiting reactant. Thus, only \(0.831\) moles of \(\text{CO}_2\) will be formed.

Step 5: Calculate the theoretical yield of \(\text{CO}_2\)

Using the moles of the limiting reactant \(\text{O}_2\), calculate the mass of \(\text{CO}_2\) produced: \[ \text{mass of } \text{CO}_2 = 0.831 \, \text{mol} \times 44.01 \, \text{g/mol} \approx 36.59 \, \text{g} \]

Key Concepts

Chemical Equations

In chemistry, a chemical equation is a symbolic representation of a chemical reaction. It shows the reactants (starting materials) and products (end materials) involved in the reaction.
In the given exercise, the chemical equation is given by: \[\text{C} + \text{O}_2 \rightarrow \text{CO}_2\]This equation illustrates that a single atom of carbon (C) combines with a molecule of oxygen (\(\text{O}_2\)) to form a molecule of carbon dioxide (\(\text{CO}_2\)).
It is essential to balance chemical equations to obey the law of conservation of mass. This law states that matter cannot be created or destroyed in a chemical reaction. In our chemical equation above, you can see that it is already balanced with one carbon and two oxygen atoms on both sides.

Molar Mass Calculation

Calculating the molar mass of each substance in your chemical equation is a crucial step in stoichiometry. Molar mass is defined as the mass of one mole of a substance (atoms, molecules, etc.) and is usually expressed in units of grams per mole (g/mol).

• The molar mass of carbon (C) is \(12.01 \, \text{g/mol}\).
• For oxygen gas (\(\text{O}_2\)), you multiply the molar mass of a single oxygen atom by two: \(2 \times 16.00 = 32.00 \, \text{g/mol}\).
• For carbon dioxide (\(\text{CO}_2\)), you add the molar mass of carbon and twice that of oxygen: \(12.01 + 2 \times 16.00 = 44.01 \, \text{g/mol}\).

Understanding molar masses allows you to convert between grams and moles, which is essential for further calculations.

Limiting Reactant

The concept of the limiting reactant is vital in chemistry as it determines how much product can be formed in a chemical reaction. It's the substance that runs out first, thus limiting the amount of product that can be generated. In our example problem, both carbon and oxygen are reactants. We calculated the moles of carbon to be approximately \(0.833\) moles and moles of \(\text{O}_2\) to be \(0.831\) moles.
According to the stoichiometry of the balanced equation, 1 mole of carbon reacts with 1 mole of \(\text{O}_2\).
Here, because the moles of \(\text{O}_2\) are slightly less than those of carbon, \(\text{O}_2\) is the limiting reactant. Consequently, it dictates the amount of \(\text{CO}_2\) that can be formed.

Theoretical Yield

Theoretical yield is the maximum amount of product that can be formed from the given amounts of reactants, assuming complete reaction with no losses. This is calculated using stoichiometry and the balanced chemical equation.In the problem, with \(0.831\) moles of \(\text{O}_2\) as our limiting reactant, we can form exactly \(0.831\) moles of \(\text{CO}_2\).
To find the mass of \(\text{CO}_2\) produced, we use its molar mass:\[\text{mass of } \text{CO}_2 = 0.831 \, \text{moles} \times 44.01 \, \text{g/mol} \approx 36.59 \, \text{g}\]Thus, the theoretical yield of carbon dioxide from this reaction is approximately 36.59 grams. Understanding theoretical yield helps in predicting the efficiency and feasibility of chemical reactions.