Potassium Nitrate Decomposition: Moles Needed For 75.2 Kg O2

Hey guys! Let's dive into a cool chemistry problem involving the decomposition of potassium nitrate ($KNO_3$). This is a classic example of stoichiometry, where we use balanced chemical equations to figure out the amounts of reactants and products involved in a reaction. Specifically, we're going to figure out how many moles of $KNO_3$ we need to heat up to get 75.2 kg of oxygen ($O_2$). Buckle up, it's gonna be a fun ride!

Understanding the Decomposition Reaction

First things first, let's take a closer look at the balanced chemical equation for the decomposition of potassium nitrate:

$4 KNO_3(s) ightarrow 2 K_2O(s) + 2 N_2(g) + 5 O_2(g)$

This equation tells us a whole bunch of stuff. It's super important, so let's break it down:

• $KNO_3(s)$: This is potassium nitrate in its solid form. It's the stuff we're starting with.
• $K_2O(s)$: This is potassium oxide, also in solid form. It's one of the products of the reaction.
• $N_2(g)$: This is nitrogen gas, another product.
• $O_2(g)$: This is oxygen gas – the stuff we're trying to produce!
• The coefficients (the big numbers in front of each chemical formula) are key. They tell us the molar ratios. For example, the '4' in front of $KNO_3$ and the '5' in front of $O_2$ means that for every 4 moles of $KNO_3$ that decompose, we get 5 moles of $O_2$. This is crucial for our calculations.

So, in essence, when we heat up potassium nitrate, it breaks down into potassium oxide, nitrogen gas, and, most importantly for us, oxygen gas. We're interested in how much potassium nitrate we need to decompose to get a specific amount of oxygen.

Calculating Moles of $O_2$

The problem states we want to produce 75.2 kg of $O_2$. But before we can use the balanced equation, we need to convert this mass into moles. Why? Because the balanced equation deals with moles, not kilograms. To do this, we'll use the molar mass of $O_2$.

Remember, the molar mass is the mass of one mole of a substance. You can find it by adding up the atomic masses of all the atoms in the molecule from the periodic table. For $O_2$, we have two oxygen atoms, and each oxygen atom has an atomic mass of approximately 16.00 g/mol. So:

Molar mass of $O_2$ = 2 * 16.00 g/mol = 32.00 g/mol

Now we can convert the mass of $O_2$ from kilograms to grams and then to moles:

75. 2 kg $O_2$ * (1000 g / 1 kg) = 75200 g $O_2$

76. 200 g $O_2$ * (1 mol $O_2$ / 32.00 g $O_2$) = 2350 moles $O_2$

So, we need to produce 2350 moles of $O_2$. We're getting closer to the answer!

Using the Stoichiometric Ratio

This is where the balanced equation comes into play again. We know from the equation that 4 moles of $KNO_3$ decompose to produce 5 moles of $O_2$. This gives us a mole ratio that we can use as a conversion factor.

The mole ratio between $KNO_3$ and $O_2$ is:

(4 moles $KNO_3$ / 5 moles $O_2$)

This ratio tells us how many moles of $KNO_3$ are required for every 5 moles of $O_2$ produced. Now we can use this ratio to find the moles of $KNO_3$ needed to produce our 2350 moles of $O_2$:

2350 moles $O_2$ * (4 moles $KNO_3$ / 5 moles $O_2$) = 1880 moles $KNO_3$

The Grand Finale: The Answer!

Therefore, we need to heat 1880 moles of $KNO_3$ to produce 75.2 kg of $O_2$. That's it! We solved the problem by carefully using the balanced chemical equation and the concept of molar ratios.

Key Takeaways and Why This Matters

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

1. Balanced Chemical Equation: We started with the balanced chemical equation, which gave us the crucial mole ratios between reactants and products.
2. Mass to Moles: We converted the given mass of $O_2$ into moles using its molar mass. This is a very common step in stoichiometry problems.
3. Stoichiometric Ratio: We used the mole ratio from the balanced equation to convert moles of $O_2$ to moles of $KNO_3$.

This type of calculation is super important in chemistry and chemical engineering. It allows us to:

• Predict product yields: We can figure out how much product we'll get from a given amount of reactant.
• Determine reactant requirements: We can figure out how much of a reactant we need to get a desired amount of product. This is essential for industrial chemical processes.
• Optimize reactions: By understanding the stoichiometry, we can optimize reaction conditions to get the best possible yield.

So, understanding stoichiometry isn't just about solving textbook problems. It's about understanding the fundamental relationships in chemical reactions, which is critical for a whole range of applications.

Practice Makes Perfect

Stoichiometry can seem a bit daunting at first, but the more you practice, the easier it becomes. Try working through similar problems, paying close attention to the units and the mole ratios. You'll be a stoichiometry pro in no time!

And that's a wrap, folks! Hope this breakdown helped you understand how to tackle this type of problem. Keep those chemistry gears turning!