Consider this equation.
2 SO2(g) + O2(g) --> 2 SO3(g)
What happens to the total volume when 3.0 L of SO2 and 2.0 L of O2 react to form the maximum amount of SO3 possible from these volumes of reactants? Assume all gases are at the same constant temperature and pressure.
***I'm not sure how to solve this problem....would ya'll mind guiding me through?
_____ SO2 (g) + _____ O2 (g) → _____ SO3 (g)
Sure, I'd be happy to help you with that! Let's break it down step by step.
First, let's determine the limiting reactant. To do this, we need to calculate the number of moles of each reactant.
For SO2:
n(SO2) = V(SO2) / Vm(SO2)
Given that V(SO2) = 3.0 L and the molar volume of SO2 (Vm(SO2)) is constant at the same temperature and pressure, we can substitute those values in to find the number of moles of SO2.
Similarly, for O2:
n(O2) = V(O2) / Vm(O2)
Given that V(O2) = 2.0 L and the molar volume of O2 (Vm(O2)) is constant at the same temperature and pressure, we can substitute those values in to find the number of moles of O2.
Since the balanced chemical equation tells us that the stoichiometric ratio between SO2 and O2 is 2:1, we can compare the number of moles of each reactant to determine which one is limiting.
Now that we know the limiting reactant, we can use the stoichiometry of the balanced equation to find the maximum amount of SO3 that can be produced.
Furthermore, we know that according to the ideal gas law, at the same temperature and pressure, the volume of a gas is directly proportional to the number of moles. So, we can use the stoichiometry of the balanced equation to determine the relationship between the number of moles and the volume of each gas.
By this relationship, we can then determine the volume of SO3 that will be produced given the number of moles and convert it back to the volume using the molar volume of SO3 (Vm(SO3)).
Lastly, to find the total volume, we need to add up the volumes of the reactant gases and the volume of the product gas.
Once we have all that information, we can determine what happens to the total volume when the reaction occurs.
I hope this guide helps you solve the problem! Let me know if you have any further questions or need more assistance.
Sure! Let's break down the problem step by step.
Step 1: Write and balance the equation
The given equation is already balanced, so we don't need to do anything in this step.
Step 2: Determine the limiting reactant
To find the limiting reactant, we need to compare the number of moles of each reactant to the stoichiometric coefficients in the balanced equation.
For SO2:
3.0 L * (1 mol / 22.4 L) = 0.134 moles
For O2:
2.0 L * (1 mol / 22.4 L) = 0.089 moles
Since the stoichiometric coefficient for SO2 is 2, and the stoichiometric coefficient for O2 is 1, we can see that O2 is the limiting reactant because it produces fewer moles of product.
Step 3: Determine the amount of product formed
The balanced equation tells us that 2 moles of SO2 react with 1 mole of O2 to produce 2 moles of SO3. Since O2 is the limiting reactant, only 1 mole of SO2 will react. Therefore, 1 mole of SO3 is produced.
Step 4: Determine the volume of product formed
To determine the volume of SO3, we need to use the ideal gas law equation: PV = nRT.
First, we need to calculate the number of moles of SO3:
1 mole / 2 = 0.5 moles
We are given the volume of reactants, but we need to find the volume at the same conditions for the product. According to the ideal gas law, the volume of a gas is directly proportional to the number of moles. Therefore, the volume of the product will be the same as the number of moles of SO3:
0.5 moles * (22.4 L / 1 mol) = 11.2 L of SO3
So, when 3.0 L of SO2 and 2.0 L of O2 react, the maximum amount of SO3 that can be formed is 11.2 L.
Of course! I'd be happy to guide you through solving this problem step by step.
To determine what happens to the total volume when the given reactants react, we need to use the coefficients of the balanced chemical equation as conversion factors.
Let's start by examining the balanced equation:
2 SO2(g) + O2(g) --> 2 SO3(g)
From this equation, we can see that for every 2 moles of SO2, we need 1 mole of O2. Also, every 2 moles of SO2 yield 2 moles of SO3. These ratios are important as they give us the conversion factors needed to solve the problem.
Now that we have the balanced equation and the given volumes, we can convert the volumes to moles using the ideal gas law equation:
PV = nRT
Given:
Volume of SO2 = 3.0 L
Volume of O2 = 2.0 L
Since all gases are at the same constant temperature and pressure, we can assume that the pressure and temperature are constant, allowing us to simplify the ideal gas law equation to:
V = nR
Next, we'll calculate the number of moles of each gas using their respective volumes and the ideal gas law equation. We'll use the molar volume of an ideal gas at constant temperature and pressure, which is 22.4 L/mol.
For SO2:
n(SO2) = V(SO2) / V(molar volume)
n(SO2) = 3.0 L / 22.4 L/mol
n(SO2) ≈ 0.134 moles
For O2:
n(O2) = V(O2) / V(molar volume)
n(O2) = 2.0 L / 22.4 L/mol
n(O2) ≈ 0.089 moles
Now we have the number of moles of each gas. It's important to note that at this point, both the reactants are in excess compared to the stoichiometric ratio required for the balanced equation. So, we can determine the limiting reagent by comparing the moles of SO2 and O2.
Since we need 2 moles of SO2 for every 1 mole of O2, it is clear that O2 is the limiting reagent because we have fewer moles of O2 (0.089 moles) compared to SO2 (0.134 moles).
Given that O2 is the limiting reagent, we can use the stoichiometry of the balanced equation to determine the maximum amount of SO3 that can be formed.
From the equation, we see that 2 moles of SO2 yield 2 moles of SO3. Therefore, 0.089 moles of O2 will yield 0.089 moles of SO3.
Now, we can use the ideal gas law equation to convert the moles of SO3 back into volume, assuming the same temperature and pressure.
V(SO3) = n(SO3) * V(molar volume)
V(SO3) = 0.089 moles * 22.4 L/mol
V(SO3) ≈ 1.99 L
Therefore, when 3.0 L of SO2 and 2.0 L of O2 react to form the maximum amount of SO3, the total volume of the gases decreases to approximately 1.99 L.