Understanding Inorganic Chemistry: Calculating Reactants in Excess for Carbon and Oxygen Reactions
Understanding Inorganic Chemistry: Calculating Reactants in Excess for Carbon and Oxygen Reactions
Chemistry, a fundamental science that deals with the composition, structure, properties, and changes of matter, has a range of theories and concepts that are crucial to our understanding of the world. One such concept is stoichiometry, which involves the quantitative relationships between reactants and products in a chemical reaction. Today, we will delve into how to calculate which reactant is in excess in a reaction involving carbon and oxygen, resulting in the formation of carbon dioxide (CO2).
Stoichiometry and Chemical Reactions
Stoichiometry is the quantitative relationship between the reactants and products in a chemical reaction. It involves the use of balanced chemical equations to relate the amounts of substances in a reaction. In the context of the problem at hand, we are dealing with the combustion of carbon with oxygen to form carbon dioxide. The balanced chemical equation for this reaction is:
2C O2 → 2CO2
This equation tells us that two moles of carbon (C) react with one mole of oxygen (O2) to produce two moles of carbon dioxide (CO2).
Quantitative Relationships in Reactants and Products
Given the problem involves 3 grams of carbon (C) reacting with 6 grams of oxygen (O2), let's proceed step-by-step to determine which reactant is in excess and how much carbon dioxide is formed.
Step 1: Calculate Moles of Each Reactant
Moles of Carbon (C): The molar mass of carbon is 12 grams per mole. Therefore, the number of moles of carbon in 3 grams is calculated as follows:moles of C 3 grams ÷ 12 grams/mole 0.25 moles
Moles of Oxygen (O2): The molar mass of oxygen (O2) is 32 grams per mole. Therefore, the number of moles of oxygen in 6 grams is calculated as follows:moles of O2 6 grams ÷ 32 grams/mole 0.1875 moles
Step 2: Determine the Limiting Reactant
The limiting reactant is the one that gets completely used up in the reaction, given the stoichiometric ratio of the reaction. Based on the balanced chemical equation, we know that 2 moles of carbon react with 1 mole of oxygen. Therefore, the ratio of carbon to oxygen required is 2:1. Let's see how many moles of oxygen would be needed to react with 0.25 moles of carbon:
moles of O2 required (0.25 moles C) × (1 mole O2 / 2 moles C) 0.125 moles O2
Since we have 0.1875 moles of oxygen, we have more than enough oxygen to react with the 0.25 moles of carbon. Therefore, oxygen is in excess, and carbon is the limiting reactant.
Step 3: Calculate Excess Reactant
To determine the amount of oxygen that remains unreacted, subtract the moles of oxygen used from the initial moles of oxygen:
moles of O2 used 0.125 moles O2 (from the limiting reactant, carbon)
moles of O2 unreacted 0.1875 moles O2 - 0.125 moles O2 0.0625 moles O2
To convert this to grams, we use the molar mass of oxygen (32 grams/mole):
mass of O2 unreacted 0.0625 moles O2 × 32 grams/mole 2 grams O2
Step 4: Calculate the Amount of CO2 Formed
Since carbon is the limiting reactant, all 0.25 moles of carbon will react, and the amount of CO2 formed will be:
moles of CO2 0.25 moles C × (2 moles CO2 / 2 moles C) 0.25 moles CO2
The molar mass of CO2 is 44 grams/mole, so the mass of CO2 formed is:
mass of CO2 0.25 moles CO2 × 44 grams/mole 11 grams CO2
In summary, the excess reactant is oxygen, and 0.75 grams of carbon remain unreacted.
Conclusion
Understanding stoichiometry is essential for predicting and explaining the behavior of chemical reactions. In the reaction between 3 grams of carbon and 6 grams of oxygen, the excess reactant is oxygen, and 0.75 grams of carbon remain unreacted, resulting in the formation of 11 grams of carbon dioxide.