HgO Decomposition: Moles Of HgO Needed For 64g Of O2

Hey guys! Let's dive into a chemistry problem involving the decomposition of mercury(II) oxide ($HgO$). This compound breaks down into mercury ($Hg$) and oxygen ($O_2$) when heated. Understanding the stoichiometry of this reaction is key to figuring out how much $HgO$ we need to produce a specific amount of oxygen. So, let's break it down step by step and get a solid understanding of the concepts involved. Let's solve this problem together, making sure we understand every single detail. The balanced chemical equation for the reaction is:

$2HgO \rightarrow 2Hg + O_2$

From this equation, we can see that 2 moles of $HgO$ decompose to produce 2 moles of $Hg$ and 1 mole of $O_2$. This mole ratio is super important for our calculations. Now, the question is: how many moles of $HgO$ are required to produce 64 grams of $O_2$? The molar mass of $O_2$ is given as 32.00 g/mol, which is a crucial piece of information.

Understanding Molar Mass and Moles

Before we jump into the calculation, let's make sure we're all on the same page about molar mass and moles. The molar mass of a substance is the mass of one mole of that substance, usually expressed in grams per mole (g/mol). For example, the molar mass of $O_2$ is 32.00 g/mol, meaning that one mole of $O_2$ weighs 32.00 grams. A mole is a unit of measurement used in chemistry to express amounts of a chemical substance. It's defined as the amount of substance containing exactly $6.02214076 × 10^{23}$ elementary entities (Avogadro's number). Moles provide a convenient way to convert between mass and number of atoms or molecules.

Converting Grams to Moles

To find out how many moles of $O_2$ are in 64 grams, we use the formula:

$\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}$

In this case:

$\text{Moles of } O_2 = \frac{64 \text{ g}}{32.00 \text{ g/mol}} = 2 \text{ moles}$

So, 64 grams of $O_2$ is equal to 2 moles of $O_2$. This conversion is the first key step in solving our problem. Understanding how to convert between grams and moles is fundamental in stoichiometry.

Stoichiometry: Using Mole Ratios

Now that we know we need to produce 2 moles of $O_2$, we can use the balanced chemical equation to determine how many moles of $HgO$ are required. The balanced equation is:

$2HgO \rightarrow 2Hg + O_2$

From the equation, we see that 2 moles of $HgO$ decompose to produce 1 mole of $O_2$. This gives us the mole ratio:

$\frac{\text{Moles of } HgO}{\text{Moles of } O_2} = \frac{2}{1}$

This ratio tells us that for every 1 mole of $O_2$ produced, we need 2 moles of $HgO$. Since we want to produce 2 moles of $O_2$, we can calculate the moles of $HgO$ needed:

$\text{Moles of } HgO = 2 \text{ moles of } O_2 × \frac{2 \text{ moles of } HgO}{1 \text{ mole of } O_2} = 4 \text{ moles of } HgO$

Therefore, we need 4 moles of $HgO$ to produce 64 grams of $O_2$. This calculation utilizes the stoichiometric coefficients from the balanced chemical equation to relate the amount of $HgO$ consumed to the amount of $O_2$ produced.

Step-by-Step Solution Summary

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

1. Convert grams of $O_2$ to moles of $O_2$: We used the molar mass of $O_2$ (32.00 g/mol) to convert 64 grams of $O_2$ to 2 moles of $O_2$.
2. Use the mole ratio from the balanced equation: We used the balanced equation $2HgO \rightarrow 2Hg + O_2$ to determine that 2 moles of $HgO$ are required to produce 1 mole of $O_2$.
3. Calculate moles of $HgO$: We multiplied the moles of $O_2$ (2 moles) by the mole ratio (2 moles $HgO$ / 1 mole $O_2$) to find that 4 moles of $HgO$ are required.

So, the final answer is 4 moles of $HgO$ are required to produce 64 grams of $O_2$.

Importance of Balanced Equations

Balanced chemical equations are absolutely crucial in stoichiometry. They provide the mole ratios needed to convert between different reactants and products in a chemical reaction. Without a balanced equation, you can't accurately determine how much of each substance is needed or produced. Always double-check that your equation is balanced before performing any calculations!

Common Mistakes to Avoid

• Forgetting to balance the equation: Always make sure the chemical equation is balanced before using it for stoichiometric calculations.
• Using incorrect molar masses: Double-check the molar masses of the substances involved. Using the wrong molar mass will lead to incorrect results.
• Incorrectly applying the mole ratio: Make sure you're using the correct mole ratio from the balanced equation. Pay attention to the coefficients in front of each substance.
• Not converting to moles first: Before using the mole ratio, make sure you have converted all masses to moles. The mole ratio relates the number of moles of reactants and products, not their masses.

Real-World Applications

Understanding stoichiometry and mole calculations isn't just for textbook problems. It has many real-world applications, including:

• Chemical Manufacturing: In industrial chemistry, stoichiometric calculations are essential for determining the correct amounts of reactants needed to produce a desired amount of product. This helps optimize production processes and minimize waste.
• Pharmaceutical Industry: Stoichiometry is used to calculate the amounts of reactants needed to synthesize drugs and other pharmaceutical compounds. Accurate calculations are crucial to ensure the purity and efficacy of the final product.
• Environmental Science: Stoichiometric calculations are used to analyze chemical reactions in the environment, such as the combustion of fuels and the formation of pollutants. This helps scientists understand and address environmental problems.
• Cooking and Baking: While it might not be obvious, cooking and baking also involve stoichiometry. Recipes are essentially stoichiometric ratios that tell you how much of each ingredient to use. Understanding these ratios can help you adjust recipes and experiment with new flavors.

Conclusion

So there you have it! We've successfully calculated the amount of $HgO$ needed to produce a specific amount of $O_2$ using stoichiometry. Remember, always start with a balanced equation, convert masses to moles, and use the mole ratio to find the desired quantity. With practice, these calculations will become second nature, and you'll be a stoichiometry master in no time! Keep practicing, and don't hesitate to ask questions. You got this! Chemistry can be challenging, but with a clear, step-by-step approach, even the most complex problems can be solved. Keep up the great work, guys!