MnO2 & O2 Production: Stoichiometry Calculation

Let's break down this stoichiometry problem step by step, guys. We need to figure out how many moles of $MnO_2$ and how many liters of $O_2$ are produced when we decompose some potassium permanganate ($KMnO_4$). There's a catch, though – our $KMnO_4$ isn't pure; it's only 80% pure, with 20% impurities. Don't worry, it's not as complicated as it sounds! We'll go through it together and make sure you understand each step. This kind of problem is classic for chemistry, and mastering it will definitely boost your confidence. First, we determine the actual mass of $KMnO_4$ that will undergo a chemical reaction. Then, using that information, determine how many moles of $MnO_2$ are produced, and finally calculate the volume of $O_2$ gas produced. This involves using the molar mass of $KMnO_4$ and the ideal gas law concept at standard temperature and pressure.

1. Calculate the Actual Mass of Pure $KMnO_4$

We are given 790 g of $KMnO_4$ which is only 80% pure. This means only 80% of the 790 g is actually $KMnO_4$. Let's calculate the mass of pure $KMnO_4$:

Mass of pure $KMnO_4$ = (Total mass) × (Purity)

Mass of pure $KMnO_4$ = 790 g × 0.80 = 632 g

So, we have 632 g of pure $KMnO_4$ that will actually react.

2. Calculate the Moles of $KMnO_4$

To figure out how much $MnO_2$ and $O_2$ are produced, we first need to know how many moles of $KMnO_4$ we're starting with. To do this, we'll use the molar mass of $KMnO_4$. The molar mass of $KMnO_4$ is calculated as follows:

• K: 39.1 g/mol
• Mn: 54.9 g/mol
• O: 16.0 g/mol (and we have 4 of them)

Molar mass of $KMnO_4$ = 39.1 + 54.9 + (4 × 16.0) = 39.1 + 54.9 + 64.0 = 158 g/mol

Now we can convert the mass of pure $KMnO_4$ to moles:

Moles of $KMnO_4$ = (Mass of pure $KMnO_4$) / (Molar mass of $KMnO_4$)

Moles of $KMnO_4$ = 632 g / 158 g/mol = 4 moles

So, we have 4 moles of $KMnO_4$.

3. Determine Moles of $MnO_2$ Produced

Now let's look at the balanced chemical equation:

$2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$

From the equation, we see that 2 moles of $KMnO_4$ decompose to produce 1 mole of $MnO_2$. We can use this stoichiometric ratio to find out how many moles of $MnO_2$ are produced from our 4 moles of $KMnO_4$. It's all about the ratios, you see. If 2 moles of $KMnO_4$ gives us 1 mole of $MnO_2$, then 4 moles of $KMnO_4$ will give us twice as much. The relationship is: 2:1. So you have half the moles of the reactant.

Moles of $MnO_2$ = (Moles of $KMnO_4$) × (Ratio of $MnO_2$ to $KMnO_4$)

Moles of $MnO_2$ = 4 moles × (1/2) = 2 moles

Therefore, 2 moles of $MnO_2$ are produced.

4. Determine Moles of $O_2$ Produced

Looking at the balanced equation again:

$2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$

We see that 2 moles of $KMnO_4$ also produce 1 mole of $O_2$. The mole ratio between $KMnO_4$ and $O_2$ is also 2:1. This means for every 2 moles of $KMnO_4$ that react, we get 1 mole of $O_2$. Same relation as $MnO_2$ which makes things easy!

Moles of $O_2$ = (Moles of $KMnO_4$) × (Ratio of $O_2$ to $KMnO_4$)

Moles of $O_2$ = 4 moles × (1/2) = 2 moles

So, we have 2 moles of $O_2$ produced.

5. Calculate the Volume of $O_2$ at STP

We need to find the volume of $O_2$ at standard temperature and pressure (STP). At STP, 1 mole of any ideal gas occupies 22.4 liters. This is a key concept to remember!

Volume of $O_2$ = (Moles of $O_2$) × (Volume of 1 mole at STP)

Volume of $O_2$ = 2 moles × 22.4 L/mole = 44.8 L

Therefore, 44.8 liters of $O_2$ are produced at STP.

6. Final Answer

We've calculated that 2 moles of $MnO_2$ and 44.8 liters of $O_2$ are produced. So, the answer is:

2 moles $MnO_2$ : 44.8 L $O_2$

This corresponds to answer choice A) 2:44.8

Key Takeaways

• Stoichiometry is Key: Understanding the mole ratios from the balanced chemical equation is crucial. Those coefficients tell you everything! Seriously, pay attention to them.
• Purity Matters: Always account for the purity of reactants. If something isn't 100% pure, you need to adjust your calculations to reflect the actual amount of the reactant present.
• STP is Your Friend: Remember that 1 mole of any gas at STP occupies 22.4 liters. This is a super handy conversion factor.

Additional Tips for Stoichiometry Problems

• Always Balance the Equation: Before you do anything else, make sure the chemical equation is balanced. An unbalanced equation will lead to incorrect mole ratios and wrong answers. Seriously, double-check this!
• Convert Everything to Moles: Moles are the language of chemistry. Convert all masses to moles before using the stoichiometric ratios.
• Pay Attention to Units: Keep track of your units throughout the calculation. This will help you avoid errors and ensure that your final answer has the correct units.
• Practice, Practice, Practice: The more stoichiometry problems you solve, the better you'll become at it. Look for practice problems in your textbook or online.

By following these steps, you can confidently tackle stoichiometry problems and ace your chemistry exams. Keep practicing, and you'll become a stoichiometry master in no time!