Chemistry: Oxygen Volume Calculation At STP

Hey there, chemistry enthusiasts! Let's dive into a classic stoichiometry problem that involves calculating the volume of oxygen gas needed to react with sulfur dioxide. We'll go through it step by step, so grab your calculators, and let's get started! This is going to be a fun and engaging way to learn how to solve this type of problem. We will be using our knowledge of chemical reactions, molar masses, and the ideal gas law. It might seem like a lot, but trust me, we will break it down into manageable chunks. By the end, you'll be a pro at calculating gas volumes!

Understanding the Chemical Reaction

First off, let's take a look at the balanced chemical equation. This equation is the foundation for our calculations. The reaction we're dealing with is the oxidation of sulfur dioxide ($SO_2$) to sulfur trioxide ($SO_3$). The balanced equation is:

$2 SO_2(g) + O_2(g) \longrightarrow 2 SO_3(g)$

This equation tells us that two moles of sulfur dioxide ($SO_2$) react with one mole of oxygen gas ($O_2$) to produce two moles of sulfur trioxide ($SO_3$). The coefficients in the balanced equation are crucial because they give us the molar ratios necessary for our calculations. The stoichiometry of the reaction is the key here, folks. It is the quantitative relationship between the reactants and products as shown by the balanced chemical equation. It essentially dictates how much of each substance we need to completely react. Think of it as a recipe where the coefficients are the amounts of each ingredient. Without knowing how the reactants and products relate to each other, we would not be able to figure out the amount of oxygen that is required to react with sulfur dioxide. In other words, the mole ratio derived from the balanced equation is critical. For example, the molar ratio of $SO_2$ to $O_2$ is 2:1, meaning that two moles of sulfur dioxide are needed to react with one mole of oxygen. This helps us to determine the amount of oxygen gas. Are you ready to proceed?

Converting Grams of $SO_2$ to Moles

Alright, our next step is to convert the given mass of sulfur dioxide ($SO_2$), which is 0.640 g, into moles. To do this, we'll need the molar mass of $SO_2$. You can find this by adding the atomic masses of sulfur (S) and two oxygen (O) atoms from the periodic table. The molar mass of sulfur is approximately 32.07 g/mol, and the molar mass of oxygen is approximately 16.00 g/mol. So, the molar mass of $SO_2$ is: (32.07 g/mol) + 2 * (16.00 g/mol) = 64.07 g/mol.

Now, we'll use the molar mass to convert the mass of $SO_2$ to moles. This is done using the following formula:

moles of $SO_2$ = (mass of $SO_2$) / (molar mass of $SO_2$)

So, moles of $SO_2$ = (0.640 g) / (64.07 g/mol) ≈ 0.00999 mol.

We have now successfully converted the mass of $SO_2$ to moles, a critical step in solving the problem. Always remember to include the units throughout your calculations to ensure accuracy! This conversion is super important because it allows us to use the balanced equation, which gives us the molar ratios, to relate the amount of $SO_2$ to the amount of $O_2$.

Using Stoichiometry to Find Moles of $O_2$

Now that we have the moles of $SO_2$, we can use the balanced equation to determine the moles of oxygen ($O_2$) that react with it. From the balanced equation ($2 SO_2(g) + O_2(g) \longrightarrow 2 SO_3(g)$), we know that 2 moles of $SO_2$ react with 1 mole of $O_2$. This gives us a mole ratio of 2:1.

To find the moles of $O_2$, we'll use the following formula:

moles of $O_2$ = (moles of $SO_2$) * (mole ratio of $O_2$ to $SO_2$)

So, moles of $O_2$ = (0.00999 mol $SO_2$) * (1 mol $O_2$ / 2 mol $SO_2$) ≈ 0.004995 mol.

This is the real deal, folks. We have successfully determined the number of moles of oxygen gas required to react with the given amount of sulfur dioxide. Remember that these ratios are derived from the balanced chemical equation. This is the cornerstone of our calculations and allows us to connect the amount of one substance to the amount of another in a chemical reaction. The magic is in the numbers. These values will pave the path for the final steps.

Calculating the Volume of $O_2$ at STP

Our final step is to calculate the volume of $O_2$ at Standard Temperature and Pressure (STP). STP is defined as a temperature of 0°C (273.15 K) and a pressure of 1 atm. At STP, one mole of any ideal gas occupies a volume of 22.4 L (or 22400 mL).

To find the volume of $O_2$ at STP, we'll use the following formula:

Volume of $O_2$ = (moles of $O_2$) * (molar volume at STP)

So, Volume of $O_2$ = (0.004995 mol) * (22400 mL/mol) ≈ 111.89 mL.

Therefore, the volume of oxygen gas required to react with 0.640 g of $SO_2$ at STP is approximately 112 mL. The use of STP allows for straightforward calculations because we know the molar volume of any gas at these conditions. Just think about it, we have taken all the steps required to determine the volume of oxygen gas at STP. How about that! We have managed to successfully navigate this problem. High five!

The Answer

Based on our calculations, the correct answer is:

C. 112 mL

Conclusion

And that's a wrap, friends! We have successfully calculated the volume of oxygen gas required to react with a given amount of sulfur dioxide at STP. We went through the balanced equation, converted grams to moles, used stoichiometry to find the moles of oxygen, and finally, calculated the volume using the molar volume at STP. Remember, practice makes perfect, so keep working on these problems to solidify your understanding. Understanding stoichiometry and the ideal gas law is super important. Until next time, happy experimenting, and keep those chemistry skills sharp!