Acetone Combustion: Calculating Air Volume Needed

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Acetone Combustion: Calculating Air Volume Needed

Hey guys! Today, we're diving into a chemistry question that involves calculating the volume of air needed for the complete combustion of acetone. This is a classic stoichiometry problem that combines chemical reactions with gas laws, so let's break it down step by step to really understand what's going on. We'll cover the balanced chemical equation, the role of oxygen in combustion, and how to calculate the air volume considering the oxygen concentration. So, buckle up and let’s get started!

Understanding Combustion and Stoichiometry

Before we jump into the specific problem, let's quickly recap the basics of combustion and stoichiometry. Combustion is a chemical process that involves the rapid reaction between a substance with an oxidant, usually oxygen, to produce heat and light. It's essentially burning something! Stoichiometry, on the other hand, deals with the quantitative relationships between reactants and products in chemical reactions. In simpler terms, it’s like a recipe for a chemical reaction, telling us exactly how much of each ingredient (reactant) we need and how much of the final product we'll get.

In a combustion reaction, a fuel (in our case, acetone) reacts with oxygen to produce carbon dioxide and water. The general form of a combustion reaction for a hydrocarbon is:

Hydrocarbon + Oxygen → Carbon Dioxide + Water

To solve stoichiometry problems, the first crucial step is to write a balanced chemical equation. This equation ensures that the number of atoms of each element is the same on both sides of the equation, adhering to the law of conservation of mass. Balancing chemical equations might seem daunting at first, but with practice, it becomes second nature. It's like learning a new language – once you grasp the grammar, you can construct any sentence! Understanding the stoichiometric coefficients (the numbers in front of the chemical formulas) is key because they tell us the molar ratios in which substances react and are produced. These coefficients are the bridge that allows us to convert between moles of reactants and products.

The Balanced Chemical Equation for Acetone Combustion

Now, let's get to the heart of our problem. Acetone has the chemical formula CH3COCH3. To figure out how much air we need for complete combustion, we first need to write the balanced chemical equation for the reaction. When acetone burns completely, it reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). Here’s the unbalanced equation:

CH3COCH3 + O2 → CO2 + H2O

Balancing this equation requires a bit of fiddling. We need to make sure that the number of carbon, hydrogen, and oxygen atoms are the same on both sides. Here’s how we can do it:

  1. Carbon: There are 3 carbon atoms in acetone (CH3COCH3), so we need 3 CO2 molecules on the product side.

    CH3COCH3 + O2 → 3 CO2 + H2O

  2. Hydrogen: There are 6 hydrogen atoms in acetone, so we need 3 H2O molecules on the product side.

    CH3COCH3 + O2 → 3 CO2 + 3 H2O

  3. Oxygen: Now, let’s count the oxygen atoms. On the product side, we have (3 × 2) + (3 × 1) = 9 oxygen atoms. On the reactant side, we have 1 oxygen atom in acetone and 2 in O2. To balance the oxygen, we need a total of 9 oxygen atoms on the reactant side. We already have 1 from acetone, so we need 8 more, which means 4 O2 molecules.

    CH3COCH3 + 4 O2 → 3 CO2 + 3 H2O

So, the balanced chemical equation for the complete combustion of acetone is:

CH3COCH3 + 4 O2 → 3 CO2 + 3 H2O

This equation is the cornerstone of our calculation. It tells us that 1 mole of acetone reacts with 4 moles of oxygen to produce 3 moles of carbon dioxide and 3 moles of water. Now that we have this balanced equation, we can use it to determine the amount of oxygen required for the combustion of 1 mole of acetone.

Calculating Oxygen Volume

The balanced equation tells us that 1 mole of acetone (CH3COCH3) requires 4 moles of oxygen (O2) for complete combustion. This is a crucial piece of information. We're given that we have 1 mole of acetone, so we know we need 4 moles of oxygen.

Now, we need to convert moles of oxygen to volume. Under standard conditions (STP: Standard Temperature and Pressure), 1 mole of any gas occupies a volume of 22.4 liters. This is a fundamental concept in chemistry and is known as the molar volume of a gas. It’s like a universal conversion factor for gases! So, to find the volume of 4 moles of oxygen, we simply multiply:

Volume of O2 = 4 moles × 22.4 L/mole = 89.6 L

So, we need 89.6 liters of pure oxygen for the complete combustion of 1 mole of acetone. But here's the catch: the problem states that the air contains 20% oxygen by volume [φ(O2) = 20%]. This means that not all of the air we're using is oxygen; only 20% of it is. This is a typical real-world scenario, as the air we breathe isn't pure oxygen—it's a mixture of gases, primarily nitrogen and oxygen, with oxygen making up about 21% of the air. The 20% in our problem is close enough for our calculations.

Accounting for Air Composition

Since air is only 20% oxygen, we need to figure out how much total air volume we need to get 89.6 liters of oxygen. To do this, we use the percentage as a conversion factor. If 20% of the air is oxygen, then the volume of oxygen is 20% of the total volume of air. Mathematically, we can express this as:

Volume of O2 = 0.20 × Total Volume of Air

We know the volume of oxygen we need (89.6 L), so we can rearrange this equation to solve for the total volume of air:

Total Volume of Air = Volume of O2 / 0.20

Plugging in our value for the volume of oxygen:

Total Volume of Air = 89.6 L / 0.20 = 448 L

Therefore, we need 448 liters of air to provide 89.6 liters of oxygen, which is the amount required for the complete combustion of 1 mole of acetone. This calculation is crucial because it reflects the real-world scenario where we're using air, not pure oxygen, for combustion. The 20% oxygen concentration significantly impacts the total volume of air needed, and failing to account for this would lead to a gross underestimation.

Final Answer and Discussion

So, the volume of air required for the complete combustion of 1 mole of acetone is 448 liters. Looking back at the multiple-choice options:

A) 448

B) 672

C) 580

D) 560

The correct answer is A) 448 liters. We've walked through the entire process, from balancing the chemical equation to accounting for the oxygen concentration in the air. This type of problem highlights the importance of understanding stoichiometry and how to apply it in practical situations. We've not just found the answer; we've also understood the chemistry behind it.

Understanding the underlying principles is key to solving similar problems in the future. For example, if we were dealing with a different fuel or a different oxygen concentration, the same approach would apply. We would start by writing the balanced chemical equation, determining the molar ratios, and then accounting for the composition of the gas mixture. These are skills that will serve you well in any chemistry endeavor.

In summary, remember these key steps for solving combustion stoichiometry problems:

  1. Write the balanced chemical equation.
  2. Determine the molar ratios between reactants.
  3. Calculate the required moles of oxygen.
  4. Convert moles of oxygen to volume using the molar volume of a gas.
  5. Account for the oxygen concentration in the air.

By following these steps, you'll be able to tackle any combustion stoichiometry problem with confidence. Keep practicing, and you'll become a pro in no time! Chemistry is all about understanding the relationships between substances and applying that knowledge to solve problems. This problem is a perfect example of how we can use stoichiometry to calculate real-world quantities. Keep exploring, keep learning, and you'll find that chemistry is not just a subject—it's a way of understanding the world around us.