Heat Released From Roasting 970g Zinc Sulfide: Calculation

Hey guys! Today, we're diving into a fascinating chemistry problem: calculating the heat released when roasting zinc sulfide. This is a crucial step in the industrial processing of zinc, and understanding the thermodynamics behind it can be super helpful. So, let's break it down step by step and make sure we've got it nailed!

Understanding the Roasting Process

Before we jump into the calculations, let's quickly chat about what roasting zinc sulfide actually means. In the industry, the first stage of processing zinc involves converting zinc sulfide (ZnS), which is the main component of zinc ore, into zinc oxide (ZnO). This is achieved by heating the ore in the presence of oxygen. The chemical equation for this process is:

2ZnS(s) + 3O₂(g) → 2ZnO(s) + 2SO₂(g) ΔH = -879 kJ

This equation tells us a lot! First, we see that two moles of solid zinc sulfide react with three moles of gaseous oxygen to produce two moles of solid zinc oxide and two moles of gaseous sulfur dioxide. The ΔH = -879 kJ is particularly important. This value represents the enthalpy change for the reaction, and the negative sign indicates that this is an exothermic reaction. In simpler terms, this means that the reaction releases heat into the surroundings – hence, the term 'roasting'. The given enthalpy change is for the reaction of 2 moles of ZnS. We will use this information to calculate the heat released for a given mass of ZnS.

Roasting is not just a simple heating process; it's a chemical transformation. The sulfur in the zinc sulfide combines with oxygen to form sulfur dioxide, a gas that is then often used in the production of sulfuric acid. Meanwhile, the zinc is converted into zinc oxide, which is a crucial intermediate in the production of metallic zinc. The heat released during the process is substantial, which is why knowing how to calculate it is so important for industrial applications. We need to be precise to ensure safety and efficiency in the industrial process, guys!

Problem Statement

Alright, now let's look at the specific problem we're tackling. We need to calculate the heat released from roasting 970 g of zinc sulfide ore. This means we need to figure out how much heat is liberated when this specific amount of ZnS reacts with oxygen. We're given the balanced chemical equation and the enthalpy change (ΔH) for the reaction, which is our starting point. The key here is to relate the given mass of ZnS to the number of moles, and then use the enthalpy change to find the heat released. We'll need to use the molar mass of ZnS to convert grams to moles. Remember, guys, stoichiometry is our friend in these kinds of calculations!

Step-by-Step Calculation

Okay, let's roll up our sleeves and get into the nitty-gritty of the calculation. Here's how we'll approach it:

Step 1: Calculate the Molar Mass of ZnS

First, we need to know the molar mass of zinc sulfide (ZnS). To do this, we'll add the atomic masses of zinc (Zn) and sulfur (S) from the periodic table.

• Atomic mass of Zn ≈ 65.38 g/mol
• Atomic mass of S ≈ 32.07 g/mol

So, the molar mass of ZnS is:

65.38 g/mol + 32.07 g/mol = 97.45 g/mol

This means that one mole of ZnS weighs approximately 97.45 grams. Remember this value; we'll use it in the next step!

Step 2: Convert Grams of ZnS to Moles

Now that we know the molar mass of ZnS, we can convert the given mass (970 g) into moles. We'll use the following formula:

Moles = Mass / Molar Mass

Plugging in the values, we get:

Moles of ZnS = 970 g / 97.45 g/mol ≈ 9.95 moles

So, we have approximately 9.95 moles of ZnS. This is a crucial step because the enthalpy change (ΔH) is given per mole of reaction. We're getting closer, guys!

Step 3: Calculate the Heat Released

Here's where the enthalpy change (ΔH) comes into play. From the balanced chemical equation, we know that the reaction of 2 moles of ZnS releases 879 kJ of heat (because ΔH = -879 kJ). We can use this information to set up a proportion and find out how much heat is released when 9.95 moles of ZnS react. The proportion looks like this:

(879 kJ) / (2 moles ZnS) = (Heat Released) / (9.95 moles ZnS)

To solve for the heat released, we'll multiply both sides by 9.95 moles:

Heat Released = (879 kJ / 2 moles ZnS) * 9.95 moles ZnS

Heat Released ≈ 4374.03 kJ

So, the heat released from roasting 970 g of zinc sulfide is approximately 4374.03 kJ. That's a significant amount of heat, guys!

Step 4: Accounting for Significant Figures and Units

It's always a good idea to think about significant figures and units. In this case, the given mass (970 g) has three significant figures, and the enthalpy change (879 kJ) also has three significant figures. Therefore, our final answer should also have three significant figures. Rounding our result to three significant figures, we get:

Heat Released ≈ 4370 kJ

So, our final answer is that approximately 4370 kJ of heat is released from roasting 970 g of zinc sulfide.

Final Answer and Summary

Okay, guys, we've crunched the numbers and arrived at our final answer! The heat released from roasting 970 g of zinc sulfide, according to the given reaction, is approximately 4370 kJ. That’s a lot of heat being generated! This calculation is a great example of how we can use stoichiometry and enthalpy changes to understand and quantify the energy involved in chemical reactions.

Let's quickly recap the steps we took:

1. Calculated the molar mass of ZnS: We found it to be approximately 97.45 g/mol.
2. Converted grams of ZnS to moles: We determined that 970 g of ZnS is approximately 9.95 moles.
3. Calculated the heat released: Using the given ΔH value and the moles of ZnS, we calculated the heat released to be approximately 4374.03 kJ.
4. Considered significant figures and units: We rounded our final answer to three significant figures, giving us 4370 kJ.

This entire process highlights the importance of understanding stoichiometry and thermochemistry in chemical reactions. By knowing the balanced equation and the enthalpy change, we can predict and calculate the heat involved in a reaction, which is crucial in many industrial processes.

Real-World Applications

Understanding the heat released during the roasting of zinc sulfide has significant real-world applications, particularly in the mining and metallurgical industries. The heat generated can be harnessed and used for other processes, increasing the overall efficiency of the industrial operation. This is a fantastic example of how chemistry principles translate into practical solutions!

Moreover, controlling the heat released is essential for safety. An uncontrolled exothermic reaction can lead to dangerous situations, so accurate calculations and careful monitoring are crucial. This knowledge helps engineers and chemists design safe and efficient processes for zinc extraction.

Conclusion

So, guys, we've successfully calculated the heat released from roasting 970 g of zinc sulfide! We navigated through the molar mass calculations, converted grams to moles, and used the enthalpy change to find our answer. This problem showcases the power of chemistry in understanding and quantifying real-world processes.

I hope this breakdown has been helpful and has made the process a bit clearer. Keep practicing these kinds of calculations, and you'll become a pro in no time! Remember, chemistry is all about understanding the interactions of matter and energy, and this problem is a perfect example of that. Keep exploring, keep learning, and have fun with chemistry!