Question

Sulfur trioxide, \(\mathrm{SO}_{3},\) is produced in enormous quantities each year for use in the synthesis of sulfuric acid. $$\begin{aligned}\mathrm{S}(s)+\mathrm{O}_{2}(g) & \longrightarrow \mathrm{SO}_{2}(g) \\\2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{SO}_{3}(g)\end{aligned}$$ What volume of \(\mathrm{O}_{2}(g)\) at \(350 .^{\circ} \mathrm{C}\) and a pressure of \(5.25\) atm is needed to completely convert \(5.00 \mathrm{g}\) sulfur to sulfur trioxide?

Short answer

Approximately 3.49 L of \(O_2(g)\) is needed to completely convert 5.00 g of sulfur to sulfur trioxide at 350 °C and 5.25 atm pressure.

Step-by-step solution

Convert mass of sulfur to moles

To convert the mass of sulfur (5.00 grams) to moles, we'll use the molar mass of sulfur. The molar mass of sulfur is 32.07 g/mol. So, to find the number of moles of sulfur, we'll divide the given mass by the molar mass: \[moles\,of\, S = \frac{5.00\,g}{32.07\,\frac{g}{mol}}\]

Calculate moles of oxygen gas needed

Now we'll use stoichiometry to determine the number of moles of oxygen gas needed by the given amount of sulfur. Examining the chemical equations, we see that for every 2 moles of \(SO_3\) formed, 1 mole of \(O_2\) is consumed. And for every 1 mole of \(SO_2\) formed, 1 mole of \(S\) is consumed. Thus: \[\frac{moles \, of \, O_2}{moles \, of \, S} = \frac{3}{2}\] Solve for moles of \(O_2\): \[moles \, of \, O_2 = \frac{3}{2} \times moles \, of \, S\]

Convert moles of oxygen gas to volume

Now we need to convert the moles of oxygen gas to volume using the Ideal Gas Law. The Ideal Gas Law states that \(PV=nRT\), where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant and T is the temperature. We are given the pressure (5.25 atm) and the temperature (350 °C or 623 K). The value of R for these units is 0.0821 \(\frac{L \cdot atm}{mol \cdot K}\). \[V = \frac{nRT}{P}\]

Calculate the volume of oxygen gas

Now we have all the necessary information to find the volume of oxygen gas. Substitute the values into the formula from Step 3: \[V = \frac{(moles\,of\, O_2) \times (0.0821\,\frac{L \cdot atm}{mol \cdot K}) \times (623 K)}{5.25 atm}\]

Combine all steps for the final solution

We'll now combine all the steps to find the volume of oxygen gas needed to completely convert 5.00 grams of sulfur to sulfur trioxide: \[V = \frac{(\frac{3}{2} \times \frac{5.00\,g}{32.07\,\frac{g}{mol}}) \times (0.0821\,\frac{L \cdot atm}{mol \cdot K}) \times (623K)}{5.25 atm}\] Calculate the volume V to obtain the final answer: \[V \approx 3.49 L\] So, 3.49 L of oxygen gas is needed to completely convert 5.00 g of sulfur to sulfur trioxide at 350 °C and 5.25 atm pressure.

Key concepts

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. By understanding the ratios of substances involved, you can predict how much of each reactant is needed to produce a desired amount of product. In our exercise, the stoichiometry of the sulfur trioxide synthesis was used to calculate the amount of oxygen needed to react with sulfur.

For every mole of sulfur consumed, the reaction requires a specific amount of oxygen gas to form sulfur dioxide, and subsequently sulfur trioxide. The balanced chemical equations provided show that one mole of sulfur reacts to form one mole of sulfur dioxide, which then reacts with half a mole of oxygen to produce sulfur trioxide. So, the stoichiometric ratio between sulfur and oxygen gas is 1:1.5. This ratio allows us to calculate how many moles of oxygen gas are necessary for the reaction given the amount of sulfur.

Ideal Gas Law

The Ideal Gas Law provides the relationship between pressure, volume, temperature, and the number of moles of a gas, which is critical in solving gas-related problems. It can be represented by the equation \( PV=nRT \), where \( P \) stands for pressure, \( V \) for volume, \( n \) for moles of gas, \( R \) for the ideal gas constant, and \( T \) for temperature in Kelvin.

In solving our exercise, after calculating the moles of oxygen gas needed using stoichiometry, the Ideal Gas Law was used to determine the volume that this quantity of gas would occupy under the given conditions of pressure and temperature. This step is essential because, under different conditions, the same amount of gas can occupy different volumes.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It is a bridge between the mass of a substance and the number of moles since it allows for conversion from grams to moles and vice versa. In our exercise, the molar mass of sulfur (32.07 g/mol) is used to find out how many moles of sulfur are present in a 5.00 gram sample.

This calculation is the first critical step in our stoichiometry problem because it provides the starting point for determining the amount of other reactants and products in the reaction, given the mass of one substance. Understanding the concept of molar mass and how to use it in calculations is a fundamental skill in chemistry.

Chemical Synthesis

Chemical synthesis involves making a desired chemical compound through a series of chemical reactions, starting from simpler substances. The synthesis of sulfur trioxide, used in the exercise, is a classic example. This process typically involves several reactions; in this case, sulfur is first burnt to form sulfur dioxide which further reacts with oxygen to yield sulfur trioxide.

This kind of synthesis is vital in industrial chemistry, as it allows for the mass production of chemicals like sulfuric acid, which is crucial in manufacturing various products. The principles of stoichiometry, molar mass, and gas laws are applied in chemical synthesis to control the quantity and purity of the final product.