Question

Explain why, in the balanced chemical equation \(\mathrm{C}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2},\) we know that \(1 \mathrm{g}\) of \(\mathrm{C}\) will not react exactly with \(1 \mathrm{g}\) of \(\mathrm{O}_{2}\).

Short answer

In the balanced chemical equation \(C + O_2 \rightarrow CO_2\), 1 gram of C does not react exactly with 1 gram of O₂ because the stoichiometry requires a 1:1 mole ratio. However, when converting the grams to moles (\(\frac{1}{12.01} \thinspace mol\) for C and \(\frac{1}{32.00} \thinspace mol\) for O₂), the actual mole ratio in this case is around 2.68, showing an excess of carbon. Therefore, O₂ is the limiting reactant and prevents the complete reaction of 1 gram of C with 1 gram of O₂.

Step-by-step solution

Determine the molar mass of reactants and products

First, we need to find the molar mass of the reactants and products in the balanced chemical equation. For carbon, the molar mass is approximately 12.01 g/mol, while for oxygen gas (O₂), the molar mass is approximately 32.00 g/mol (as each oxygen atom has a molar mass of approximately 16.00 g/mol). The molar mass of carbon dioxide (CO₂) is the sum of the molar masses of carbon and oxygen, which is approximately 44.01 g/mol.

Determine the mole-to-mole ratio of reactants

The balanced chemical equation already shows the stoichiometric coefficients for each reactant and product: C + O₂ → CO₂. This means that 1 mole of carbon reacts with 1 mole of oxygen gas to produce 1 mole of carbon dioxide.

Convert grams to moles for the reactants

Now we need to convert the given amount of each reactant (1 gram) to moles: For carbon: \( \frac{1 \thinspace g}{12.01 \thinspace g/mol} = 0.083 \thinspace mol \) For oxygen gas: \( \frac{1 \thinspace g}{32.00 \thinspace g/mol} = 0.031 \thinspace mol \)

Compare the mole ratio to the stoichiometric ratio of the reactants

Comparing the mole ratio of the given amounts of carbon and oxygen gas (0.083 mol / 0.031 mol ≈ 2.68) to the stoichiometric ratio (1:1), we can see that there is more carbon available to react with the available oxygen gas. This means that oxygen gas (O₂) is the limiting reactant in this case, while there is an excess of carbon. Therefore, 1 gram of carbon will not react exactly with 1 gram of oxygen gas to produce carbon dioxide. In conclusion, based on the mole-to-mole ratio of the balanced chemical equation and the available amount of reactants, it is clear that 1 gram of carbon does not react exactly with 1 gram of oxygen gas to produce carbon dioxide.

Key concepts

Molar Mass

To understand why 1 gram of carbon does not react exactly with 1 gram of oxygen gas, we must first grasp the concept of molar mass. Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). Each element has a unique molar mass based on its atomic weight. For example, the molar mass of carbon (C) is approximately 12.01 g/mol. This means that one mole of carbon atoms weighs approximately 12.01 grams.

Oxygen (O), being a diatomic molecule in its natural state (O₂), has a molar mass of 32.00 g/mol. The reason behind this value is that each oxygen atom has a molar mass of about 16.00 g/mol, so two oxygen atoms in O₂ collectively have a molar mass of 32.00 g/mol. Understanding molar mass is crucial when converting between grams and moles, which is a fundamental step in stoichiometry.

By knowing the molar mass of each reactant, we convert the mass of substances into moles to compare them on an equal footing during chemical reactions.

Mole-to-Mole Ratio

In stoichiometric calculations, the mole-to-mole ratio is the proportional relationship between the amounts, in moles, of the reactants and products involved in a chemical reaction. This ratio is derived directly from the balanced chemical equation, which provides the stoichiometric coefficients. For our reaction of carbon and oxygen to produce carbon dioxide:

C + O₂ → CO₂;

the equation tells us that 1 mole of carbon reacts with 1 mole of oxygen to form 1 mole of carbon dioxide. This 1:1:1 ratio is the mole-to-mole ratio.

A balanced equation's mole-to-mole ratio allows us to predict how many moles of products will be formed from a given amount of reactants, ensuring that mass and atom balance are maintained through the reaction. This understanding is crucial for accurately determining the amounts of reactants needed and the expected yield of products.

Limiting Reactant

The limiting reactant in a chemical reaction is the substance that is completely consumed first, thereby limiting the extent of the reaction and determining the amount of product formed. In our given problem, we observe that when 1 gram each of carbon and oxygen gas are used, a comparison of the mole quantities becomes necessary.

Using the molar mass, we find that 1 gram of carbon corresponds to approximately 0.083 moles, while 1 gram of oxygen gas corresponds to about 0.031 moles. According to the 1:1 mole ratio from the balanced equation, oxygen is insufficient to completely react with the available carbon.

This deficiency means oxygen is the limiting reactant, as it will run out first, halting the production of carbon dioxide. The excess carbon remains unreacted because no more oxygen is available to continue the reaction. Identifying the limiting reactant is a crucial step in predicting the quantities of products obtained from a given reaction.

Balanced Chemical Equation

A balanced chemical equation is essential for accurately representing chemical reactions. It ensures that the same number of each type of atom is present on both the reactant and product sides. This is in accordance with the Law of Conservation of Mass, which states that mass is neither created nor destroyed in a chemical reaction.

The balanced equation for the reaction of carbon and oxygen is:

C + O₂ → CO₂.

This equation signifies that one atom of carbon reacts with one molecule of oxygen (containing two atoms) to form one molecule of carbon dioxide, maintaining a balance of all atoms involved.

Balancing chemical equations involves adjusting the stoichiometric coefficients to ensure equal numbers of each element's atoms appear on both sides of the equation. This process is fundamental to performing calculations related to reaction stoichiometry, as it provides the mole-to-mole ratios necessary for determining reactant consumption and product formation.