Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation for this is
S2 (g) + C (s) <---> CS2 (g)
Kc = 9.40 at 900 K
How many grams of CS2 (g) can be prepared by heating 11.7 moles of S2 (g) with excess carbon in a 5.95 L reaction vessel at 900 K until equilibrium is reached?
M S2 = mols/L = ?
.......C + S2 ==> CS2
I.....xs..?M.......0
C..........-x......x
E.........?M-x.....x
Substitute the E line into the Kc expression and solve for x (in mols/L)
Then mols = M x L = ?
Then g CS2 = mols CS2 x molar mass CS2
To solve this problem, you need to use the given information and apply the principles of equilibrium and stoichiometry.
Step 1: Write the balanced chemical equation:
S2 (g) + C (s) ↔ CS2 (g)
Step 2: Determine the molar ratio between S2 and CS2:
From the balanced equation, the molar ratio is 1:1. This means that for every mole of S2 reacted, one mole of CS2 is produced.
Step 3: Calculate the number of moles of CS2 that can be prepared:
Since the molar ratio between S2 and CS2 is 1:1, the number of moles of CS2 that can be produced is also 11.7 moles.
Step 4: Convert moles of CS2 to grams:
To convert moles to grams, you need to use the molar mass of CS2. The molar mass of CS2 is:
(12.01 g/mol × 1) + (32.07 g/mol × 2) = 76.15 g/mol
So, the mass of CS2 that can be prepared is:
Mass = Moles × Molar mass
Mass = 11.7 moles × 76.15 g/mol = 891.02 g
Step 5: Considering the given reaction vessel volume:
The volume of the reaction vessel is given as 5.95 L.
Step 6: Interpretation of equilibrium constant (Kc):
The equilibrium constant (Kc) is given as 9.40.
Kc = [CS2] / [S2] = 9.40
Step 7: Calculate the concentration at equilibrium:
To calculate the concentration at equilibrium, you need to convert the mass of CS2 to moles and divide it by the volume of the reaction vessel.
Concentration = Moles / Volume
Concentration = 891.02 g / (5.95 L × 76.15 g/mol) = 2.64 M
Step 8: Use the equilibrium constant to solve for the unknown concentration:
Kc = [CS2] / [S2]
9.40 = [2.64 M] / [11.7 M]
Rearrange the equation to solve for [CS2]:
[CS2] = Kc × [S2]
[CS2] = 9.40 × [11.7 M]
[CS2] = 109.98 M
Step 9: Calculate the mass of CS2 at equilibrium:
Mass = Concentration at equilibrium × Molar mass
Mass = 109.98 M × 76.15 g/mol = 8366.47 g
However, since we are starting with 891.02 g of S2, the maximum amount of CS2 that can be produced is limited by the amount of S2 available. Therefore, the maximum amount of CS2 that can be prepared is equal to 891.02 g.
So, the maximum amount of CS2 that can be prepared by heating 11.7 moles of S2 with excess carbon in a 5.95 L reaction vessel at 900 K until equilibrium is reached is 891.02 grams.
To find out how many grams of CS2 can be prepared in this reaction, you need to use the given equilibrium constant (Kc) and the information about the moles of S2 and the volume of the reaction vessel.
Here's how you can solve it step by step:
Step 1: Write down the balanced chemical equation:
S2 (g) + C (s) <--> CS2 (g)
Step 2: Calculate the moles of CS2 formed at equilibrium using the given moles of S2.
In the balanced equation, the stoichiometric ratio between S2 and CS2 is 1:1. This means that for every 1 mole of S2, 1 mole of CS2 is produced. Therefore, if you have 11.7 moles of S2, you will also have 11.7 moles of CS2 at equilibrium.
Step 3: Convert the moles of CS2 to grams.
To convert moles to grams, you need to use the molar mass of CS2. The molar mass of carbon disulfide (CS2) is 76.14 g/mol. Multiply the moles of CS2 by its molar mass:
11.7 moles CS2 x 76.14 g/mol = 891.18 grams of CS2
Therefore, you can prepare approximately 891.18 grams of CS2 by heating 11.7 moles of S2 with excess carbon in a 5.95 L reaction vessel at 900 K until equilibrium is reached.