Calcium Nitrate Decomposition: A Chemistry Problem Solved

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Hey guys! Today, we're diving deep into a classic chemistry problem involving the decomposition of calcium nitrate. We'll break down the equation, balance it like pros, calculate the relative formula mass, and even analyze a decomposition scenario. Let's get started and make chemistry a little less intimidating, shall we?

Balancing the Chemical Equation for Calcium Nitrate Decomposition

The first step in tackling any stoichiometry problem is to have a balanced chemical equation. We're given the unbalanced equation for the decomposition of calcium nitrate:

Ca(NO₃)₂ → CaO + NO₂ + O₂

Our mission, should we choose to accept it (and we do!), is to make sure that the number of atoms for each element is the same on both sides of the equation. This is crucial because it reflects the law of conservation of mass, which basically says that matter can't be created or destroyed in a chemical reaction.

So, how do we balance this equation? Let's take a systematic approach. We'll start by counting the number of atoms for each element on both sides:

  • Reactants (Left Side):
    • Calcium (Ca): 1
    • Nitrogen (N): 2
    • Oxygen (O): 6
  • Products (Right Side):
    • Calcium (Ca): 1
    • Nitrogen (N): 1
    • Oxygen (O): 5

Notice that calcium is already balanced, which is a nice little win for us! However, nitrogen and oxygen are out of whack. Nitrogen has 2 atoms on the left and only 1 on the right. Oxygen has 6 on the left and 5 on the right. We need to find coefficients (the numbers we put in front of the chemical formulas) that will balance these elements.

A good strategy is to start with the element that appears in the fewest compounds. In this case, let's focus on nitrogen first. To balance nitrogen, we can put a coefficient of 2 in front of NOâ‚‚:

Ca(NO₃)₂ → CaO + 2NO₂ + O₂

Now, let's recount the atoms:

  • Reactants (Left Side):
    • Calcium (Ca): 1
    • Nitrogen (N): 2
    • Oxygen (O): 6
  • Products (Right Side):
    • Calcium (Ca): 1
    • Nitrogen (N): 2
    • Oxygen (O): 7

Nitrogen is balanced! But now oxygen is even more unbalanced. We have 6 on the left and 7 on the right. This is where it gets a little tricky, because oxygen appears in all three compounds on the product side. When you have an element in multiple compounds, it's often best to leave it for last.

Instead of directly balancing oxygen, let's look at the odd number of oxygen atoms on the product side (7). A common trick for dealing with odd numbers of atoms is to try and make them even. We can do this by putting a coefficient of 2 in front of Ca(NO₃)₂:

2Ca(NO₃)₂ → CaO + 2NO₂ + O₂

But wait! This messes up our calcium balance. We now have 2 calcium atoms on the left and only 1 on the right. No worries, we can fix this by putting a coefficient of 2 in front of CaO:

2Ca(NO₃)₂ → 2CaO + 2NO₂ + O₂

Let's recount those atoms again:

  • Reactants (Left Side):
    • Calcium (Ca): 2
    • Nitrogen (N): 4
    • Oxygen (O): 12
  • Products (Right Side):
    • Calcium (Ca): 2
    • Nitrogen (N): 2
    • Oxygen (O): 7

Okay, calcium is balanced. But now we have 4 nitrogen atoms on the left and only 2 on the right. We can fix this by changing the coefficient in front of NOâ‚‚ to 4:

2Ca(NO₃)₂ → 2CaO + 4NO₂ + O₂

Recount time!

  • Reactants (Left Side):
    • Calcium (Ca): 2
    • Nitrogen (N): 4
    • Oxygen (O): 12
  • Products (Right Side):
    • Calcium (Ca): 2
    • Nitrogen (N): 4
    • Oxygen (O): 10

Nitrogen is balanced, but oxygen is still off. We have 12 on the left and only 10 on the right. We're getting closer! The oxygen atoms in Oâ‚‚ are the only ones we can adjust at this point. We need 2 more oxygen atoms on the product side to reach 12. Since each Oâ‚‚ molecule has 2 oxygen atoms, we can put a coefficient of 2 in front of Oâ‚‚:

2Ca(NO₃)₂ → 2CaO + 4NO₂ + 2O₂

One final recount:

  • Reactants (Left Side):
    • Calcium (Ca): 2
    • Nitrogen (N): 4
    • Oxygen (O): 12
  • Products (Right Side):
    • Calcium (Ca): 2
    • Nitrogen (N): 4
    • Oxygen (O): 12

Boom! We did it! The equation is balanced. Our balanced equation is:

2Ca(NO₃)₂ → 2CaO + 4NO₂ + 2O₂

Balancing chemical equations can feel like solving a puzzle, but with practice, you'll become a master at it. Remember to take it step-by-step and double-check your work!

Calculating the Relative Formula Mass of Calcium Nitrate

Next up, let's tackle the relative formula mass (RFM) of calcium nitrate, Ca(NO₃)₂. The relative formula mass is the sum of the atomic masses of all the atoms in a formula unit. It's a crucial concept for converting between mass and moles, which we'll need later when we analyze the decomposition.

To calculate the RFM, we need the relative atomic masses (RAM) of each element in the compound. You can usually find these on a periodic table. Here are the RAMs we'll need:

  • Calcium (Ca): 40.1
  • Nitrogen (N): 14.0
  • Oxygen (O): 16.0

Now, let's break down the formula Ca(NO₃)₂. We have:

  • 1 Calcium (Ca) atom
  • 2 Nitrogen (N) atoms (because of the subscript 2 outside the parentheses)
  • 6 Oxygen (O) atoms (3 oxygen atoms inside the parentheses, multiplied by the 2 outside)

To calculate the RFM, we multiply the RAM of each element by the number of atoms of that element and then add them all together:

RFM of Ca(NO₃)₂ = (1 × RAM of Ca) + (2 × RAM of N) + (6 × RAM of O)

RFM of Ca(NO₃)₂ = (1 × 40.1) + (2 × 14.0) + (6 × 16.0)

RFM of Ca(NO₃)₂ = 40.1 + 28.0 + 96.0

RFM of Ca(NO₃)₂ = 164.1

So, the relative formula mass of calcium nitrate is 164.1. Remember, RFM doesn't have any units, but it's often expressed in grams per mole (g/mol) when we're dealing with molar mass.

Calculating the RFM might seem tedious at first, but it's a fundamental skill in chemistry. With practice, you'll be able to calculate it in your sleep (well, maybe not, but you'll get good at it!).

Decomposition Analysis: From Grams to Moles and Beyond

Now for the final piece of our chemistry puzzle! We're told that 16.4 g of calcium nitrate is decomposed, and we're going to analyze this decomposition using our balanced equation and RFM. This is where the real magic of stoichiometry happens – we're going to use the relationships between reactants and products to figure out how much of each substance is involved.

The first thing we need to do is convert the mass of calcium nitrate (16.4 g) into moles. Moles are the chemist's favorite unit for measuring amounts of substances because they directly relate to the number of particles (atoms, molecules, etc.). We can use the following formula to convert grams to moles:

Moles = Mass / Molar Mass

We already calculated the molar mass of calcium nitrate (which is numerically equal to the RFM) as 164.1 g/mol. So, let's plug in the values:

Moles of Ca(NO₃)₂ = 16.4 g / 164.1 g/mol

Moles of Ca(NO₃)₂ ≈ 0.1 mol

So, we have approximately 0.1 moles of calcium nitrate decomposing. Now, we can use the balanced equation to figure out how many moles of the other products are formed.

Remember our balanced equation:

2Ca(NO₃)₂ → 2CaO + 4NO₂ + 2O₂

The coefficients in the balanced equation tell us the mole ratios between the reactants and products. For example:

  • 2 moles of Ca(NO₃)â‚‚ decompose to produce 2 moles of CaO
  • 2 moles of Ca(NO₃)â‚‚ decompose to produce 4 moles of NOâ‚‚
  • 2 moles of Ca(NO₃)â‚‚ decompose to produce 2 moles of Oâ‚‚

We can use these mole ratios to calculate the moles of each product formed from 0.1 moles of Ca(NO₃)₂. Let's start with calcium oxide (CaO):

From the balanced equation, the mole ratio of Ca(NO₃)₂ to CaO is 2:2, which simplifies to 1:1. This means that for every 1 mole of Ca(NO₃)₂ that decomposes, 1 mole of CaO is formed. So, if we start with 0.1 moles of Ca(NO₃)₂, we'll form 0.1 moles of CaO.

Next, let's look at nitrogen dioxide (NOâ‚‚):

The mole ratio of Ca(NO₃)₂ to NO₂ is 2:4, which simplifies to 1:2. This means that for every 1 mole of Ca(NO₃)₂ that decomposes, 2 moles of NO₂ are formed. So, if we start with 0.1 moles of Ca(NO₃)₂, we'll form 0.1 mol * 2 = 0.2 moles of NO₂.

Finally, let's consider oxygen (Oâ‚‚):

The mole ratio of Ca(NO₃)₂ to O₂ is 2:2, which simplifies to 1:1. This means that for every 1 mole of Ca(NO₃)₂ that decomposes, 1 mole of O₂ is formed. So, if we start with 0.1 moles of Ca(NO₃)₂, we'll form 0.1 moles of O₂.

So, in summary, when 16.4 g (0.1 moles) of calcium nitrate decomposes, it produces:

    1. 1 moles of Calcium Oxide (CaO)
    1. 2 moles of Nitrogen Dioxide (NOâ‚‚)
    1. 1 moles of Oxygen (Oâ‚‚)

We can even convert these moles back into grams if we wanted to, using the same formula we used earlier (Mass = Moles × Molar Mass). But for now, we've successfully analyzed the decomposition and figured out the amounts of each product formed.

Analyzing decomposition reactions like this is a fundamental skill in chemistry. It allows us to predict the outcomes of chemical reactions and to understand the quantitative relationships between reactants and products.

Final Thoughts

So, guys, we've tackled a complete chemistry problem, from balancing the equation to analyzing the decomposition. We've seen how the balanced equation and relative formula mass are essential tools for understanding chemical reactions. Chemistry can be challenging, but by breaking down problems step-by-step and practicing regularly, you'll become more confident and successful. Keep up the great work, and never stop exploring the fascinating world of chemistry! Remember, it's all about understanding the relationships and applying the concepts. You've got this!