Calculating Oxygen Consumption In Coal Combustion A Stoichiometry Problem

Hey guys! Have you ever stopped to think about how much oxygen is consumed when a large amount of coal is burned? It's a fascinating question, especially when we bring in concepts from chemistry and physics. Today, let's dive deep into calculating the volume of oxygen consumed in the complete combustion of coal, a topic that is not only intriguing but also highly relevant for exams like the ENEM (Exame Nacional do Ensino Médio) and other entrance exams.

Understanding the Basics of Coal Combustion

Before we jump into the calculations, let’s make sure we’re all on the same page with the basics.

What is Combustion?

Combustion, at its core, is a chemical process that involves the rapid reaction between a substance with an oxidant, usually oxygen, to produce heat and light. In simpler terms, it's what happens when something burns. Think of lighting a match or the burning of fuel in a car engine – these are everyday examples of combustion.

The Role of Coal in Combustion

Coal, a fossil fuel, is primarily composed of carbon. When coal burns, the carbon in it reacts with oxygen in the air. This reaction is exothermic, meaning it releases energy in the form of heat. This heat is what we harness in power plants to generate electricity.

The Chemical Equation

The chemical equation for the complete combustion of carbon is relatively straightforward:

${ C + O_2 \rightarrow CO_2 }$

This equation tells us that one atom of carbon (C) reacts with one molecule of oxygen (${O_2}$) to produce one molecule of carbon dioxide (${CO_2}$). This simple equation is the foundation for our calculations.

Key Concepts: Molar Mass and CNTP

To perform our calculations accurately, we need to understand a couple of key concepts:

• Molar Mass: The molar mass of a substance is the mass of one mole of that substance. The molar mass of carbon (C) is approximately 12 g/mol, and the molar mass of oxygen (${O_2}$) is approximately 32 g/mol.
• CNTP (Standard Temperature and Pressure): CNTP conditions are defined as 0°C (273.15 K) and 1 atm pressure. Under these conditions, one mole of any gas occupies a volume of 22.4 liters. This is a crucial piece of information for our calculations.

Practical Implications

Understanding coal combustion is not just an academic exercise. It has significant practical implications:

• Energy Production: Coal is a major source of energy worldwide. Understanding its combustion helps us optimize energy production processes.
• Environmental Impact: The combustion of coal releases carbon dioxide, a greenhouse gas, into the atmosphere. Understanding the stoichiometry of the reaction helps us quantify and mitigate the environmental impact.

By grasping these fundamental concepts, we set the stage for tackling the problem at hand: determining the volume of oxygen consumed in the combustion of a large quantity of coal. So, let’s put on our thinking caps and get ready to crunch some numbers!

Problem Statement and Initial Data

Okay, guys, now that we’ve covered the basics, let’s tackle the specific problem we’re trying to solve. We have a scenario where a large amount of coal is burned daily, and we need to figure out how much oxygen is consumed in the process. This involves some stoichiometry and a bit of gas law knowledge, but don’t worry, we’ll break it down step by step. So, let's dive into understanding the problem statement and the initial data we have.

The Problem Statement

Here’s the scenario we’re working with: An industry burns 1,200 kg of coal (carbon) daily, with the coal being 90% pure. We're assuming that the combustion is complete, meaning all the carbon reacts with oxygen to form carbon dioxide. The big question we need to answer is: What volume of oxygen is consumed in this combustion process under CNTP (Standard Temperature and Pressure) conditions?

This problem is a classic example of a stoichiometry question, which involves calculating the amounts of reactants and products in a chemical reaction. In this case, our reactants are carbon and oxygen, and our product is carbon dioxide. We need to use the balanced chemical equation for the combustion of carbon, which we discussed earlier, and the molar masses of the substances involved to find our answer.

Gathering the Initial Data

To solve this problem, we need to organize the information we have. Let’s list the key data points:

• Mass of Coal Burned Daily: 1,200 kg
• Purity of Coal: 90% carbon
• Reaction: Complete combustion (${ C + O_2 \rightarrow CO_2 }$)
• Conditions: CNTP (Standard Temperature and Pressure)

From this data, we can derive some additional information that will be useful in our calculations:

• Molar Mass of Carbon (C): Approximately 12 g/mol
• Molar Mass of Oxygen (${O_2}$): Approximately 32 g/mol
• Molar Volume of a Gas at CNTP: 22.4 liters/mol

Converting Units

Before we start calculating, it’s essential to ensure our units are consistent. We have the mass of coal in kilograms, but we’ll need to work with grams to align with the molar mass units. So, let’s convert the mass of coal from kilograms to grams:

${ 1,200 \text{ kg} = 1,200 \times 1,000 \text{ g} = 1,200,000 \text{ g} }$

Now we have the mass of coal in grams, which is compatible with the molar mass units. The next step is to account for the purity of the coal.

Accounting for Coal Purity

The coal is only 90% pure carbon, which means that not all of the 1,200,000 grams is carbon. We need to calculate the mass of pure carbon in the coal:

${ \text{Mass of pure carbon} = 1,200,000 \text{ g} \times 0.90 = 1,080,000 \text{ g} }$

So, we have 1,080,000 grams of pure carbon that will react with oxygen. Now we’re ready to move on to the stoichiometric calculations.

Setting Up the Problem

With the initial data in hand and the units sorted out, we’ve laid a solid foundation for solving the problem. We know the mass of pure carbon burned, and we have the chemical equation for the combustion reaction. The next step is to use this information to calculate the amount of oxygen consumed. So, let’s move on to the next section where we'll perform the stoichiometric calculations to find the answer!

Step-by-Step Calculation of Oxygen Volume

Alright, let’s get into the nitty-gritty of the calculation! Now that we know how much pure carbon is being burned, we can figure out how much oxygen is needed for this reaction. Remember, stoichiometry is our friend here – it's all about the ratios in the balanced chemical equation. So, let's break down the calculation step by step to make sure we don’t miss anything.

Step 1: Convert Mass of Carbon to Moles

First things first, we need to convert the mass of carbon to moles. Why? Because chemical reactions happen in mole ratios, not mass ratios. We know the mass of pure carbon burned is 1,080,000 grams, and the molar mass of carbon is approximately 12 g/mol. So, we use the formula:

${ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} }$

Plugging in our values:

${ \text{Moles of Carbon} = \frac{1,080,000 \text{ g}}{12 \text{ g/mol}} = 90,000 \text{ moles} }$

So, we have 90,000 moles of carbon being burned. That’s a lot of carbon! But we’re not done yet. Now we need to figure out how much oxygen this will consume.

Step 2: Determine Moles of Oxygen Required

This is where the balanced chemical equation comes into play. Remember, the equation for the combustion of carbon is:

${ C + O_2 \rightarrow CO_2 }$

This equation tells us that 1 mole of carbon reacts with 1 mole of oxygen (${O_2}$). The mole ratio between carbon and oxygen is 1:1. This makes our job a bit easier because the number of moles of carbon burned will be equal to the number of moles of oxygen consumed.

So, if we have 90,000 moles of carbon, we’ll need 90,000 moles of oxygen to react with it completely.

${ \text{Moles of Oxygen} = 90,000 \text{ moles} }$

Great! We now know how many moles of oxygen are required. The final step is to convert this to a volume under CNTP conditions.

Step 3: Convert Moles of Oxygen to Volume at CNTP

Under CNTP conditions, 1 mole of any gas occupies 22.4 liters. This is a crucial piece of information for this type of problem. We can use this to convert the moles of oxygen to volume using the following formula:

${ \text{Volume} = \text{Moles} \times \text{Molar Volume at CNTP} }$

Plugging in our values:

${ \text{Volume of Oxygen} = 90,000 \text{ moles} \times 22.4 \text{ liters/mol} }$

${ \text{Volume of Oxygen} = 2,016,000 \text{ liters} }$

So, the volume of oxygen consumed in the combustion of 1,200 kg of coal (90% pure) is 2,016,000 liters under CNTP conditions. That’s a massive amount of oxygen!

Summarizing the Calculation

Let’s recap the steps we took to solve this problem:

1. Converted the mass of coal to grams.
2. Accounted for the purity of the coal to find the mass of pure carbon.
3. Converted the mass of carbon to moles.
4. Used the balanced chemical equation to determine the moles of oxygen required.
5. Converted the moles of oxygen to volume at CNTP.

By following these steps, we were able to systematically solve the problem and find the volume of oxygen consumed. Now, let's put our final answer in context and see why this calculation matters.

Final Answer and its Significance

Okay, guys, we’ve done the hard work and crunched the numbers. We found that the volume of oxygen consumed in the complete combustion of 1,200 kg of coal (90% pure) is 2,016,000 liters under CNTP conditions. That’s a pretty significant amount, isn’t it? But what does this number really mean, and why should we care?

The Final Answer

So, to reiterate, our final answer is:

${ \text{Volume of Oxygen Consumed} = 2,016,000 \text{ liters} }$

This means that every day, the industry in our scenario consumes over two million liters of oxygen just to burn coal. That's enough oxygen to fill several Olympic-sized swimming pools! Putting the number in this perspective helps us appreciate the scale of the combustion process.

Significance of the Calculation

Understanding the oxygen consumption in combustion processes is crucial for several reasons:

• Industrial Applications: Industries that rely on combustion, such as power plants, need to ensure they have an adequate supply of oxygen for efficient operation. If oxygen supply is limited, combustion may be incomplete, leading to reduced energy output and increased emissions of pollutants like carbon monoxide.
• Environmental Impact: The combustion of fossil fuels like coal has a significant impact on the environment. It releases large amounts of carbon dioxide, a major greenhouse gas, into the atmosphere. By understanding the stoichiometry of the reaction and the amount of oxygen consumed, we can better estimate the amount of carbon dioxide produced. This information is vital for developing strategies to mitigate climate change.
• Safety Considerations: Incomplete combustion can also lead to the formation of hazardous substances. Ensuring sufficient oxygen supply is essential for safe combustion processes, especially in enclosed environments.

Connecting to Real-World Scenarios

Let’s think about some real-world scenarios where this type of calculation is important:

• Power Plants: Coal-fired power plants use combustion to generate electricity. They need to calculate the amount of oxygen required for efficient operation and to minimize emissions.
• Industrial Furnaces: Many industrial processes, such as steelmaking and cement production, use furnaces that burn fuel. Calculating oxygen consumption helps optimize these processes.
• Emergency Response: In the event of a fire, understanding oxygen consumption can help firefighters assess the situation and determine the best course of action.

Implications for Exams Like ENEM

This type of problem is also highly relevant for exams like the ENEM. These exams often include questions that test your understanding of stoichiometry, gas laws, and the environmental impact of chemical processes. Being able to solve problems like this demonstrates a strong grasp of these concepts.

Final Thoughts

So, guys, we've come to the end of our journey through calculating oxygen consumption in coal combustion. We've seen how a seemingly simple chemical reaction can have significant implications for industry, the environment, and even our performance on exams. By breaking down the problem step by step and understanding the underlying principles, we've shown how to tackle complex calculations with confidence.

Remember, the key to mastering these types of problems is practice and a solid understanding of the basics. Keep practicing, keep asking questions, and you’ll be well on your way to acing your exams and making a positive impact on the world around you! Keep rocking it!