For 2 SO2(g) + O2(g) 2 SO3(g), Kp = 3.0 104 at 700 K. In a 2.00-L vessel the equilibrium mixture contains 1.58 g of SO3 and 0.136 g of O2. How many grams of SO2 are in the vessel?
There are a couple of ways to approach this. The simpler approach may be to change Kp to Kc, convert grams SO3 and grams O2 to M, set up an ICE(you need only the E line) chart and substitute into the Kc expression, solve for (SO2) then change to grams.
(SO3) = mols/L = 1.58/2*80 = ?
(O2) = mols/L = 0.136/2*32
........2SO2 + O2 ==> 2SO3
E........x.0.00213....0.00988
The second approach is to use Kp as is. Convert grams SO3 and grams O2 to mols and plug into PV = nRT; solve for pO2 and pSO3 separately. Then construct an ICE chart (you need only the E line) and substitute into Kp expression and solve for pSO2. Plug that back into PV = nRT to solve for n and convert to grams.
Well, to solve this problem, we need a chemical equation and some math skills. But let me tell you a joke first to lighten the mood:
Why don't scientists trust atoms? Because they make up everything!
Now, let's get back to the problem. We have the equation 2 SO2(g) + O2(g) ⇌ 2 SO3(g) and we want to find the mass of SO2 in the vessel.
First, we need to calculate the number of moles of SO3 and O2 using their respective masses. The molar masses of SO3, O2, and SO2 are approximately 80 g/mol, 32 g/mol, and 64 g/mol, respectively.
So, the number of moles of SO3 is given by:
moles of SO3 = mass of SO3 / molar mass of SO3
= 1.58 g / 80 g/mol
≈ 0.01975 mol
Similarly, the number of moles of O2 is:
moles of O2 = mass of O2 / molar mass of O2
= 0.136 g / 32 g/mol
≈ 0.00425 mol
Now, let's use the stoichiometry of the equation to determine the number of moles of SO2. According to the equation, 2 moles of SO2 react with 1 mole of O2 and give 2 moles of SO3.
Since we know the moles of O2 and SO3, we can use the following ratio:
moles of SO2 = (2 moles of SO2 / 1 mole of O2) × moles of O2
= 2 × 0.00425 mol
≈ 0.0085 mol
Finally, we can calculate the mass of SO2 using its molar mass:
mass of SO2 = moles of SO2 × molar mass of SO2
= 0.0085 mol × 64 g/mol
≈ 0.544 g
So, approximately 0.544 grams of SO2 are in the vessel.
To find the number of grams of SO2 in the vessel, we need to use the given equilibrium constant (Kp) and the amounts of SO3 and O2 present.
1. First, convert the masses of SO3 and O2 to moles using their molar masses:
- The molar mass of SO3 (sulfur trioxide) is 80.06 g/mol, so the number of moles of SO3 is:
Number of moles of SO3 = mass of SO3 / molar mass of SO3 = 1.58 g / 80.06 g/mol
- The molar mass of O2 (oxygen) is 32.00 g/mol, so the number of moles of O2 is:
Number of moles of O2 = mass of O2 / molar mass of O2 = 0.136 g / 32.00 g/mol
2. Now, let's find the number of moles of SO2 (sulfur dioxide) using the balanced equation:
The stoichiometry of the reaction tells us that the ratio of moles of SO2 to moles of O2 is 2:1. Since O2 is already given, we need to find the moles of O2, which we calculated in step 1.
Moles of SO2 = Moles of O2 × (2/1)
3. Convert the moles of SO2 to grams using its molar mass:
Mass of SO2 = Moles of SO2 × molar mass of SO2
Now, let's perform the calculations to find the grams of SO2 in the vessel:
1. Number of moles of SO3:
Number of moles of SO3 = 1.58 g / 80.06 g/mol = 0.0197 mol
2. Number of moles of O2:
Number of moles of O2 = 0.136 g / 32.00 g/mol = 0.00425 mol
3. Number of moles of SO2:
Number of moles of SO2 = 0.00425 mol × (2/1) = 0.00850 mol
4. Mass of SO2:
Mass of SO2 = 0.00850 mol × 64.07 g/mol = 0.545 g
Therefore, there are 0.545 grams of SO2 in the vessel.
To find the number of grams of SO2 in the vessel, we need to use the given information and apply the concept of stoichiometry.
First, let's determine the moles of SO3 and O2 in the vessel:
Moles of SO3 = mass of SO3 / molar mass of SO3
Moles of O2 = mass of O2 / molar mass of O2
The molar masses of SO3 and O2 can be found using the periodic table. The molar mass of SO3 is approximately 80.06 g/mol, and the molar mass of O2 is approximately 32.00 g/mol.
Moles of SO3 = 1.58 g / 80.06 g/mol
Moles of O2 = 0.136 g / 32.00 g/mol
Next, we can use the balanced equation, 2 SO2(g) + O2(g) -> 2 SO3(g), to establish the mole ratio between SO2 and SO3.
From the equation, we can see that 2 moles of SO2 react to produce 2 moles of SO3. This means that for every mole of SO3, there must have been one mole of SO2 originally.
Since the moles of SO3 represent the moles of SO2 in the equilibrium mixture, we can conclude that the number of moles of SO2 is the same as the number of moles of SO3 present.
Finally, we can convert the moles of SO2 to grams:
Mass of SO2 = moles of SO2 x molar mass of SO2
The molar mass of SO2 is approximately 64.06 g/mol.
Therefore, we can calculate the mass of SO2 in the vessel using the equation:
Mass of SO2 = moles of SO3 x molar mass of SO2
Note: The molar mass of SO2 was not provided, so I used the most common value. If you have a different molar mass, you can substitute it into the equation.
I hope this explanation helps you understand how to find the grams of SO2 in the vessel!