Question

When \(0.100 \mathrm{~mol}\) of carbon is burned in a closed vessel with \(8.00 \mathrm{~g}\) of oxygen, how many grams of carbon dioxide can form? Which reactant is in excess, and how many grams of it remain after the reaction?

Short answer

4.401 grams of carbon dioxide are formed. Oxygen is in excess with 4.80 grams remaining.

Step-by-step solution

Write the balanced chemical equation

The reaction of carbon with oxygen to form carbon dioxide can be described by the balanced chemical equation: \[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \]

Calculate the moles of oxygen

Given mass of oxygen is 8.00 g. Use the molar mass of oxygen (\text{O}_2), which is 32.00 g/mol, to find moles. \[ \text{Moles of } \text{O}_2 = \frac{8.00 \text{ g}}{32.00 \text{ g/mol}} = 0.250 \text{ mol} \]

Determine the limiting reactant

From the balanced equation, 1 mole of carbon reacts with 1 mole of oxygen. We have 0.100 moles of carbon and 0.250 moles of oxygen. Since there is less carbon than oxygen and a 1:1 ratio is required, carbon is the limiting reactant.

Calculate the moles of carbon dioxide formed

According to the balanced equation, 1 mole of carbon produces 1 mole of carbon dioxide. As carbon is the limiting reactant and we have 0.100 moles of it, the same amount of carbon dioxide is produced. \[ \text{Moles of } \text{CO}_2 = 0.100 \text{ mol} \]

Convert moles of carbon dioxide to grams

Use the molar mass of carbon dioxide (\text{CO}_2), which is 44.01 g/mol, to find the mass. \[ \text{Mass of } \text{CO}_2 = 0.100 \text{ mol} \times 44.01 \text{ g/mol} = 4.401 \text{ g} \]

Determine the amount of excess reactant

Since carbon is the limiting reactant, we need to calculate how much oxygen remains. Initially, there are 0.250 moles of oxygen. The reaction consumes 0.100 moles of oxygen (same as the moles of carbon). \[ \text{Moles of excess } \text{O}_2 = 0.250 \text{ mol} - 0.100 \text{ mol} = 0.150 \text{ mol} \]

Convert moles of excess oxygen to grams

Use the molar mass of oxygen to convert the excess oxygen back to grams. \[ \text{Mass of excess } \text{O}_2 = 0.150 \text{ mol} \times 32.00 \text{ g/mol} = 4.80 \text{ g} \]

Key concepts

limiting reactant

A limiting reactant is the reactant in a chemical reaction that runs out first, stopping the reaction from continuing because there is no more left to react. In our exercise, we have carbon and oxygen reacting to form carbon dioxide. We use the balanced chemical equation: \[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \]Start by calculating the moles of each reactant. For carbon, we are given 0.100 moles. For oxygen, we find the moles by converting the given mass (8.00 g) to moles using its molar mass (32.00 g/mol), leading to 0.250 moles of oxygen. Because the reaction requires a 1:1 ratio of carbon to oxygen, carbon is the limiting reactant as we only have 0.100 moles of it compared to 0.250 moles of oxygen. The reaction stops when all the carbon is used up.

moles to grams conversion

Moles to grams conversion is an essential step in stoichiometry, as it allows us to understand how much mass of a product is created or how much of a reactant is needed. In our exercise, we need to convert the moles of reactants and products to grams.

For example, to find out how many grams of carbon dioxide (\text{CO}_2) are produced, we first note that 0.100 moles of \text{C} produce the same number of moles of \text{CO}_2. We then use the molar mass of \text{CO}_2, which is 44.01 g/mol:

\[ \text{Mass of } \text{CO}_2 = 0.100 \text{ mol} \times 44.01 \text{ g/mol} = 4.401 \text{ g} \]

This calculation shows us that 4.401 grams of carbon dioxide will form.

balanced chemical equations

Balanced chemical equations are crucial because they provide the exact ratio of reactants and products in a chemical reaction. It ensures that matter is conserved and the calculations based on these ratios are accurate.

In our exercise, the balanced equation is: \[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \]This indicates that one mole of carbon reacts with one mole of oxygen to produce one mole of carbon dioxide. Balancing an equation involves ensuring that the number of atoms of each element is the same on both sides of the equation, which follows the Law of Conservation of Mass. Without a balanced equation, it would be impossible to correctly identify the limiting reactant or accurately calculate the amounts of products formed.

excess reactant

The excess reactant is the substance that remains after a chemical reaction has completed. It is the reactant that is not entirely consumed when the reaction stops. Identifying the excess reactant can help in understanding resource efficiency and further potential reactions.

In the given exercise, we identified carbon as the limiting reactant. Therefore, oxygen is in excess. Initially, there are 0.250 moles of \text{O}_2. The reaction consumes 0.100 moles of \text{O}_2 (equal to the moles of carbon).

\[ \text{Moles of excess } \text{O}_2 = 0.250 \text{ mol} - 0.100 \text{ mol} = 0.150 \text{ mol} \]We convert these moles back to grams using the molar mass of oxygen (32.00 g/mol):

\[ \text{Mass of excess } \text{O}_2 = 0.150 \text{ mol} \times 32.00 \text{ g/mol} = 4.80 \text{ g} \]

This calculation shows that 4.80 grams of oxygen are left unreacted after the reaction completes.