Aluminum Nitrate Decomposition: Calculate Oxygen Mass

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Hey guys! Let's dive into a fascinating chemistry problem involving the decomposition of solid aluminum nitrate upon heating. This is a classic type of problem often encountered in chemistry, and understanding the steps involved will really boost your problem-solving skills. We'll break down the question, walk through the calculations, and make sure you grasp the core concepts. So, let's get started and figure out how to calculate the mass of oxygen formed when aluminum nitrate decomposes. Understanding chemical reactions and stoichiometry is key to solving this problem.

Understanding the Problem

First, let’s clearly understand what the problem is asking. We're told that solid aluminum nitrate (Al(NO3)3) is heated. When heated, aluminum nitrate decomposes, meaning it breaks down into simpler substances. The mass of the solid decreases from 12.78 g to 6.48 g. This mass loss is due to the release of gases, one of which is oxygen (O2). Our mission is to calculate the mass of oxygen that is produced during this reaction. To tackle this, we will utilize the principles of mass conservation and stoichiometry. Stoichiometry, in simple terms, deals with the quantitative relationships between reactants and products in chemical reactions. It’s like a recipe for a chemical reaction, telling us how much of each ingredient (reactant) we need and how much of the final product we will get.

Before we jump into calculations, it's crucial to visualize what's happening. Aluminum nitrate decomposition is a chemical change where a single compound breaks down into multiple products. The general balanced equation for the decomposition of metal nitrates typically involves the formation of a metal oxide, nitrogen dioxide, and oxygen gas. We need to figure out the exact balanced equation for the decomposition of aluminum nitrate to solve this problem accurately. This balanced equation will give us the molar ratios between aluminum nitrate and oxygen, which are essential for our calculations. Once we have the balanced equation, we can use the change in mass to determine the amount of oxygen produced.

Writing the Balanced Chemical Equation

The cornerstone of solving any stoichiometry problem is the balanced chemical equation. It tells us the exact ratio of reactants and products involved in the reaction. For the decomposition of aluminum nitrate, the unbalanced equation looks something like this:

Al(NO3)3(s) → Al2O3(s) + NO2(g) + O2(g)

See how aluminum nitrate (Al(NO3)3) breaks down into aluminum oxide (Al2O3), nitrogen dioxide (NO2), and oxygen (O2)? Now, we need to balance this equation to ensure that the number of atoms of each element is the same on both sides of the arrow. Balancing chemical equations is a skill that comes with practice, but here’s how we can do it for this reaction:

  1. Balance Aluminum (Al): There are 2 Al atoms on the product side (Al2O3), so we need 2 Al(NO3)3 on the reactant side: 2Al(NO3)3(s) → Al2O3(s) + NO2(g) + O2(g)
  2. Balance Nitrogen (N): Now we have 6 nitrogen atoms (2 * 3) on the reactant side, so we need 6 NO2 molecules on the product side: 2Al(NO3)3(s) → Al2O3(s) + 6NO2(g) + O2(g)
  3. Balance Oxygen (O): On the reactant side, we have 18 oxygen atoms (2 * 3 * 3). On the product side, we have 3 (from Al2O3) + 12 (from 6NO2) + 2 (from O2) = 17 oxygen atoms. To balance this, we need to add 1/2 O2 to the product side: 2Al(NO3)3(s) → Al2O3(s) + 6NO2(g) + 1/2O2(g)
  4. Get Rid of Fractions: To get rid of the fraction, we multiply the entire equation by 2: 4Al(NO3)3(s) → 2Al2O3(s) + 12NO2(g) + 3O2(g)

So, the balanced chemical equation is:

4Al(NO3)3(s) → 2Al2O3(s) + 12NO2(g) + 3O2(g)

This equation tells us that 4 moles of aluminum nitrate decompose to produce 2 moles of aluminum oxide, 12 moles of nitrogen dioxide, and 3 moles of oxygen gas. We’ll use this crucial information to calculate the mass of oxygen formed.

Calculating the Mass of Oxygen Formed

Now that we have our balanced chemical equation, we can use the information given in the problem to calculate the mass of oxygen produced. Remember, the mass decreased by 6.3 g during the reaction. This mass loss represents the gases that were released – nitrogen dioxide (NO2) and oxygen (O2). To find out how much of that mass loss is due to oxygen, we’ll use stoichiometry.

First, let’s determine the change in mass. The problem states that the mass decreased from 12.78 g to 6.48 g. The difference is:

Change in mass = 12.78 g - 6.48 g = 6.30 g

This 6.30 g represents the combined mass of the gases released, which are nitrogen dioxide and oxygen. To find the mass of oxygen specifically, we need to work with moles. We'll first find the molar mass of aluminum nitrate and then relate it to the molar mass of oxygen using the balanced equation.

The molar mass of Al(NO3)3 can be calculated by adding the atomic masses of each element in the compound:

  • Aluminum (Al): 26.98 g/mol
  • Nitrogen (N): 14.01 g/mol
  • Oxygen (O): 16.00 g/mol

Molar mass of Al(NO3)3 = 26.98 + 3(14.01 + 3(16.00)) = 26.98 + 3(14.01 + 48.00) = 26.98 + 3(62.01) = 26.98 + 186.03 = 213.01 g/mol

Next, we need the molar mass of oxygen (O2). Since the atomic mass of oxygen is 16.00 g/mol, the molar mass of O2 is 2 * 16.00 = 32.00 g/mol.

Now, let's use the balanced equation (4Al(NO3)3(s) → 2Al2O3(s) + 12NO2(g) + 3O2(g)) to set up a proportion. From the balanced equation, we see that 4 moles of Al(NO3)3 produce 3 moles of O2. We can use this ratio to convert the mass loss to moles of oxygen.

However, we need to first figure out how many moles of Al(NO3)3 were initially present. We don't have this information directly from the mass loss, but we can relate the total mass loss to the moles of oxygen produced through a series of steps involving the molar masses and the stoichiometric coefficients from the balanced equation.

The key insight here is that the mass loss corresponds to the gases released, and we need to find the portion of that mass loss that is due to oxygen. We will use the mole ratio between the total moles of gases produced and the moles of oxygen produced to figure this out.

To simplify the calculation, let's first think about the ratio of moles of O2 produced to the total moles of gases (NO2 and O2) produced. From the balanced equation, 3 moles of O2 and 12 moles of NO2 are produced for every 4 moles of Al(NO3)3 decomposed. So, for every 3 moles of O2, there are 12 moles of NO2, making a total of 15 moles of gas produced.

So the ratio of moles of O2 to total moles of gas is 3/15 = 1/5.

Now, we know that the mass loss of 6.30 g is the combined mass of NO2 and O2. Let's assume we have 'x' moles of O2. Then, we have 5x moles of total gases (NO2 and O2). So, the moles of NO2 would be 5x - x = 4x.

We can now set up an equation using the molar masses of O2 and NO2:

(x mol O2 * 32.00 g/mol) + (4x mol NO2 * 46.01 g/mol) = 6.30 g

32x + 184.04x = 6.30

216.04x = 6.30

x = 6.30 / 216.04

x ≈ 0.02916 mol O2

Finally, we can calculate the mass of oxygen formed by multiplying the moles of O2 by its molar mass:

Mass of O2 = 0.02916 mol * 32.00 g/mol

Mass of O2 ≈ 0.933 g

Final Answer

Therefore, approximately 0.933 grams of oxygen were formed during the decomposition of aluminum nitrate. This problem highlights the importance of understanding and applying the concepts of stoichiometry, balanced chemical equations, and molar masses. It might seem a bit complex at first, but breaking it down step-by-step makes it much more manageable. You got this!

Key Takeaways

Let's recap the key concepts we used to solve this problem:

  • Balanced Chemical Equations: The cornerstone of stoichiometry. Make sure you can confidently balance equations before tackling these types of problems.
  • Molar Mass: Knowing how to calculate molar mass is crucial for converting between grams and moles.
  • Stoichiometric Ratios: Using the coefficients in the balanced equation to establish mole ratios between reactants and products.
  • Mass Conservation: The total mass of the reactants equals the total mass of the products.

By mastering these concepts, you'll be well-equipped to handle a wide range of stoichiometry problems in chemistry. Keep practicing, and you'll become a pro in no time! Remember, chemistry is like building with LEGOs – each concept builds upon the previous one. So, make sure your foundations are strong, and you'll be able to construct even the most complex solutions. Keep up the great work, guys!