Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is S2 +C =CS2
Kc= 9.40 at 900k
How many grams of CS2(g) can be prepared by heating 14.3 moles of S2(g) with excess carbon in a 8.55 L reaction vessel held at 900 K until equilibrium is attained?
.022
............S2(g) + C(s) ==> CS2(g)
I..........1.67.....--.......0
C...........-x.......--......x
E.........1.67-x....--.......x
Note: (S2) = 14.3mols/8.55L = 1.6725 M.
Kc = 9.40 = (CS2)/(S2)
Substitute the equilibrium line into Kc expression and solve for x = (CS2) in mols/L.
x mols/L x 8.55 L = mols CS2
g CS2 = mols x molar mass.
Oh, chemistry, you sneaky little science! Let's get down to business and solve this riddle.
First things first, we need to find the number of moles of CS2 that can be prepared. To do this, we need to use the balanced equation: S2 + C = CS2.
From the equation, we can see that the stoichiometric ratio between S2 and CS2 is 1:1. So, if we have 14.3 moles of S2, we can expect to form the same number of moles of CS2.
Now, to find the mass of CS2, we need to multiply the number of moles by its molar mass. The molar mass of CS2 is 76.13 g/mol.
So, the mass of CS2 can be calculated as follows:
Mass = Moles × Molar Mass
Mass of CS2 = 14.3 moles × 76.13 g/mol
I'll leave the math to you, my friend. Just remember, no clowning around when it comes to calculations!
To find the grams of CS2(g) that can be prepared, we first need to calculate the number of moles of CS2 using the given values and then convert it to grams.
Step 1: Convert moles of S2 to moles of CS2
According to the balanced chemical equation, 1 mole of S2 reacts to produce 1 mole of CS2.
So, the number of moles of CS2 is also 14.3 moles.
Step 2: Convert moles of CS2 to grams of CS2
The molar mass of CS2 can be calculated by adding the atomic masses of carbon (C) and sulfur (S) in the molecule.
C = 12.01 g/mol
S = 32.07 g/mol
So, the molar mass of CS2 is:
Molar mass of CS2 = (12.01 g/mol + 32.07 g/mol) = 44.08 g/mol
To find the grams of CS2(g), we can multiply the number of moles of CS2 by its molar mass:
Grams of CS2(g) = 14.3 moles * 44.08 g/mol = 630.44 grams
Therefore, 630.44 grams of CS2(g) can be prepared by heating 14.3 moles of S2(g) with excess carbon in a 8.55 L reaction vessel held at 900 K until equilibrium is attained.
To answer this question, we need to use the given information and the equilibrium constant (Kc) to determine the amount of CS2 that can be prepared. We'll follow these steps:
Step 1: Write the balanced chemical equation:
S2 + C = CS2
Step 2: Calculate the moles of S2:
The given information states that we have 14.3 moles of S2.
Step 3: Use the stoichiometry of the balanced equation to determine the moles of CS2 produced:
From the balanced equation, we can see that 1 mole of S2 reacts to form 1 mole of CS2. Therefore, we will have 14.3 moles of CS2.
Step 4: Convert the moles of CS2 to grams:
To do this, we need to know the molar mass of CS2, which is 76.14 g/mol. Multiply the moles of CS2 by its molar mass to get the mass in grams:
Mass of CS2 = 14.3 moles × 76.14 g/mol = 1088.062 g
Step 5: Determine the volume of the reaction vessel:
The given information states that the reaction vessel has a volume of 8.55 L.
Step 6: Use the ideal gas law to check if conditions are favorable for the reaction:
The ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. Given the volume (V = 8.55 L) and temperature (T = 900 K), we can check if the pressure is within acceptable limits or if conditions need to be adjusted in order to attain equilibrium.
Once these steps are followed, you can determine that 1088.062 grams of CS2 can be prepared by heating 14.3 moles of S2 with excess carbon in an 8.55 L reaction vessel held at 900 K until equilibrium is attained.