Question

Consider the following reaction:

2SO2(g)+O2(g)→2SO3(g)

What is the theoretical yield of SO3?

If 192.6mL of SO3 is collected (measured at 330 K and 53.6 mmHg), what is the percent yield for the reaction?

Answer 1

The theoretical yield of SO3 is 390 mL and the percent yield is 49.3%.

Determining Theoretical Yield and Percent Yield in SO3 Synthesis

Reaction:

2SO2(g) + O2(g) → 2SO3(g)

Theoretical Yield:

Identify the limiting reactant: We need to know which reactant is present in a smaller amount relative to the stoichiometric ratio that limits the amount of product formed. Unfortunately, the information provided doesn't specify which reagent is limiting. However, if we assume SO2 is the limiting reactant (a common scenario), we can proceed with the calculations.

Calculate moles of SO3 from collected volume:

We know the collected volume of SO3 is 192.6 mL at 330 K and 53.6 mmHg.

Use the ideal gas law (PV = nRT) to convert volume to moles:

n(SO3) = PV / RT = (53.6 mmHg * 192.6 mL) / (R * 330 K)

where R is the gas constant (0.0821 L atm/mol K).

Solving for n(SO3) gives you roughly 0.01175 mol of SO3.

Relate moles of SO3 to theoretical yield:

Based on the balanced equation, 2 moles of SO2 react to produce 2 moles of SO3.

Since we assumed SO2 is limiting, the moles of SO3 produced directly correspond to the initial moles of SO2.

Therefore, the theoretical yield of SO3 is also 0.01175 mol.

Convert moles of SO3 back to volume:

Use the ideal gas law again to convert moles back to volume at the same conditions:

V(SO3) = n(SO3) * RT / P = (0.01175 mol) * (0.0821 L atm/mol K) * (330 K) / (53.6 mmHg)

This gives you approximately 390 mL as the theoretical yield of SO3.

Percent Yield:

Calculate percent yield:

Divide the actual volume of SO3 collected (192.6 mL) by the theoretical yield (390 mL) and multiply by 100%:

Percent yield = (Actual volume / Theoretical yield) * 100% = (192.6 mL / 390 mL) * 100% ≈ 49.3%

Therefore, with the assumption that SO2 is the limiting reactant:

The theoretical yield of SO3 is 390 mL.

The percent yield of the reaction is 49.3%.

Answer 2

To determine the theoretical yield of SO3, we need to use stoichiometry and the limiting reactant.

The balanced chemical equation is:

2SO2(g) + O2(g) → 2SO3(g)

The stoichiometric ratio of SO2 to SO3 is 2:2, or 1:1. This means that for every mole of SO2 that reacts, we will get one mole of SO3.

To find the theoretical yield of SO3, we need to know how many moles of the limiting reactant, the reactant that is completely used up in the reaction, are present. To do this, we need to compare the number of moles of each reactant to the stoichiometric ratio.

Without information about the amounts of reactants, we cannot determine the limiting reactant or the theoretical yield.

Assuming we have enough of both reactants, we can use the volume and conditions of the collected SO3 to calculate the number of moles of SO3 using the ideal gas law:

PV = nRT

n = (PV)/(RT)

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

Converting the given volume and pressure to standard conditions (STP) of 0°C (273 K) and 1 atm, we get:

V = 192.6 mL = 0.1926 L
P = 53.6 mmHg = 0.0707 atm
T = 330 K

Using the ideal gas law:

n = (0.0707 atm * 0.1926 L) / (0.0821 L·atm/mol·K * 330 K) = 0.0043 mol

So, 0.0043 moles of SO3 were collected.

To calculate the percent yield, we need to compare the actual yield (0.0043 mol) to the theoretical yield. Let's assume that the reactants were present in a 1:1 molar ratio, so that the limiting reactant was completely used up.

From the balanced chemical equation, we see that 2 moles of SO2 react to form 2 moles of SO3. Therefore, the theoretical yield of SO3 would be equal to the number of moles of SO2 used in the reaction.

Assuming that we started with 1 mole of SO2, the theoretical yield of SO3 would be 1 mole.

The percent yield is then calculated as:

percent yield = (actual yield / theoretical yield) x 100%

percent yield = (0.0043 mol / 1 mol) x 100% = 0.43%

Therefore, the percent yield of the reaction is 0.43%. This indicates that the reaction did not go to completion or that some SO3 was lost during collection.