2 SO2 + O2 --> 2 SO3

What happens to the total volume when 3L of SO2 react with 2L of O2 to form the max. amount of SO3 possible?

To determine what happens to the total volume when 3L of SO2 reacts with 2L of O2 to form the maximum amount of SO3 possible, we need to examine the balanced chemical equation.

The balanced equation is:
2 SO2 + O2 → 2 SO3

From this equation, we can see that for every 2 moles of SO2 reacting, we need 1 mole of O2. This corresponds to a volume ratio of 2:1.

Given that 3L of SO2 is reacting, we can calculate the corresponding volume of O2 needed:
3L SO2 × (1L O2 / 2L SO2) = 1.5L O2

Since we only have 2L of O2 available, which is less than the required amount of 1.5L, the limiting reactant is O2. The amount of SO3 formed will be determined by the amount of O2 available.

Now, let's determine the maximum amount of SO3 formed. Since the stoichiometric ratio is 2:1, for every 2 moles of SO2 reacting, we form 2 moles of SO3. This means that the moles of SO3 formed will be equal to the moles of O2 available.

Using the ideal gas equation, we can assume that the volume of the reaction is directly proportional to the number of moles of gas involved. Therefore, the total volume of the reactants and products will be the sum of the volumes of the individual gases.

In this case, the total volume will be equal to the sum of the volumes of SO2 and SO3.

Total volume = volume of SO2 + volume of SO3

Since we have 3L of SO2 and 2 moles of SO3 formed from 2 moles of O2, the volume of SO3 formed will be the same as the volume of O2 used, which is 2L.

Therefore, the total volume when 3L of SO2 reacts with 2L of O2 to form the maximum amount of SO3 possible will be:
Total volume = 3L SO2 + 2L SO3 = 3L + 2L = 5L.

To determine what happens to the total volume when the given reaction occurs, you need to first calculate the number of moles of each reactant and product. Then, use the stoichiometry of the reaction to determine the coefficient relationship between the reactants and products. Finally, apply the ideal gas law to determine the volume.

Let's start by calculating the number of moles for each reactant and product:
- SO2: 3 L
- O2: 2 L

Since we know the volumes, we can assume the conditions are constant, such as temperature and pressure. Therefore, we can directly compare the volumes of the gases based on their number of moles.

Next, we need to consider the balanced equation:
2 SO2 + O2 -> 2 SO3

According to the stoichiometry, 2 moles of SO2 react with 1 mole of O2 to produce 2 moles of SO3. Therefore, the ratio of reactants to products is 2:1.

Now let's calculate the number of moles for the products:
- SO2: 3 moles
- O2: 2 moles

Since the ratio is 2:1, the limiting reactant in this case is O2 because it is available in fewer moles. This means that 2 moles of SO3 will be produced based on 2 moles of O2 being consumed.

Finally, compare the initial moles of SO2 and O2 to the moles of the product SO3. In this case, since we have an equal amount of SO2 and O2 in moles (3 moles each), we will have an equal volume of SO3.

Therefore, in the given reaction, the total volume remains the same when 3L of SO2 react with 2L of O2 to form the maximum amount of SO3 possible.

How much SO3 can form if all of the SO2 is used?

How much SO3 can form if all of the O2 is used?
So which is the limiting reagent and which reactant will be in excess?\

The correct answer is that ????? L SO3 will be formed and ????L of ???? (the limiting reagent) will remain.

You might be screwed, very difficult question but the answer might be C