S2 + C ----> CS2
How many grams of CS2 can be prepared by heating 11.0 moles of S2 with excess carbon in 5.20 L reaction vessel held at 900 kelvin until equilibrium is attained?
Kc= 9.40
(S2) = 11.0mols/5.20L = approx 2 but you need to be more accurate.
.........S2 + C ==> CS2
I mols...2..xs...0
C.......-x..xs...x
E......2-x..xs...x
Kc = (CS2)/(S2) note:C doesn't appear in the Kc exprssion because it is a solid.
Substitute the E line into Kc expression and solve for x = (CS2) = M in mols/L.
total mols = M x L = ?
grams = mols x molar mass = ?
To solve this problem, we need to use the equilibrium constant (Kc) and the stoichiometry of the balanced chemical equation to determine the amount of CS2 that can be prepared.
The balanced chemical equation for the reaction is:
S2 + C ----> CS2
From the balanced equation, we can see that the stoichiometric ratio between S2 and CS2 is 1:1.
We can use the ideal gas law to convert the volume of the reaction vessel to the number of moles of reactants. The ideal gas law is given by:
PV = nRT
Where:
P = Pressure of the gas (assumed constant)
V = Volume of the gas
n = Number of moles of gas
R = Ideal gas constant
T = Temperature in Kelvin
In this case, the volume of the reaction vessel is 5.20 L, the pressure is assumed constant, and the temperature is 900 Kelvin. We can rearrange the ideal gas law equation to solve for the number of moles:
n = PV / RT
Substituting the values:
n = (5.20 L) / (0.0821 L * atm / mol * K * 900 K)
n = 0.2578 moles
Since we have 11.0 moles of S2 and the stoichiometric ratio is 1:1, we can conclude that S2 is in excess.
Now let's calculate the number of moles of CS2 using the equilibrium constant (Kc) and the number of moles of S2:
Kc = [CS2] / [S2]
Kc = (moles of CS2) / (0.2578 moles)
Solving for the moles of CS2:
(moles of CS2) = Kc * 0.2578 moles
(moles of CS2) = 9.40 * 0.2578 moles ≈ 2.42 moles
Finally, we can convert the moles of CS2 to grams using the molar mass of CS2, which is approximately 76.14 g/mol:
mass of CS2 = (moles of CS2) * (molar mass of CS2)
mass of CS2 ≈ 2.42 moles * 76.14 g/mol ≈ 184.48 grams
Therefore, approximately 184.48 grams of CS2 can be prepared by heating 11.0 moles of S2 with excess carbon in a 5.20 L reaction vessel held at 900 Kelvin until equilibrium is attained.
To determine the number of grams of CS2 that can be prepared, you need to use the given information and the principles of stoichiometry.
1. Write down the balanced chemical equation:
S2 + C ----> CS2
2. Find the molar mass of CS2:
The molar mass of carbon disulfide (CS2) is the sum of the atomic masses of carbon (C) and sulfur (S):
C: 12.01 g/mol
S: 32.07 g/mol
CS2: 12.01 g/mol + (2 * 32.07 g/mol) = 76.14 g/mol
3. Calculate the number of moles of CS2 using the given moles of S2:
Using stoichiometry, we know that the ratio of moles of S2 to moles of CS2 is 1:1. Therefore, the number of moles of CS2 will also be 11.0 moles.
4. Calculate the volume of CS2 produced:
The volume of CS2 is determined by the volume of the reaction vessel. In this case, the volume is 5.20 L.
5. Calculate the number of moles of CS2 using the volume and temperature:
To do this, we'll need to use the ideal gas law. The ideal gas law is given by the equation:
PV = nRT
Where:
P = pressure (assuming constant)
V = volume
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature in Kelvin
Given:
V = 5.20 L
T = 900 K
R = 0.0821 L.atm/mol.K
Now, let's rearrange the ideal gas equation to solve for n (the number of moles):
n = PV / RT
Substituting the given values:
n = (5.20 L * P) / (0.0821 L.atm/mol.K * 900 K)
6. Use the value of n to calculate the mass of CS2:
The mass of CS2 will be the number of moles (n) multiplied by the molar mass of CS2:
mass = n * molar mass
mass = 11.0 moles * 76.14 g/mol
Calculate the mass to get the answer.