Answer:
974.6 grams of CS2 can be prepared.
Explanation:
<u>Step 1:</u> Data given
Kc = 9.40 at 900 K
Moles of S2 = 14.2 mol
volume = 6.30 L
Molar mass of CS2 = 76.14 g/mol
<u>Step 2:</u> The balanced equation:
S2(g)+C(s) ↔ CS2(g)
<u>Step 3:</u> Calculate initial concentrations
Concentration of S2 = moles S2 / volume
Concentration of S2 = 14.2 moles / 6.30 L
Concentration of S2 = 2.25 M (This is the initial concentration)
The initial concentration of C and CS2 is 0M
Since the mole ratio is 1:1:1
There will react X
The concentration of S2 at the equilibrium is: (2.25 -X)M
The concentration of C and CS2 at the equilibrium is X M
<u>Step 4:</u> Calculate concentrations
Since C is not a gas but solid, it doesn't matter for the Kc
Kc = 9.40 = (products)/(reactants) = [CS2(g)]/[S2(g)]
9.40 = X/(2.25-X)
X = 2.034 = [CS2]
[S2] = 2.25 - 2.034 = 0.216
<u>Step 5</u>: Calculate moles of CS2
Moles CS2 = molarity CS2 * volume
Moles CS2 = 2.034 M * 6.30 L
Moles CS2 = 12.8 moles
<u>Step 6:</u> Calculate mass of CS2
Mass CS2 = moles CS2 * molar mass CS2
Mass CS2 = 12.8 moles * 76.14 g/mol
Mass CS2 = 974.6 grams
974.6 grams of CS2 can be prepared.
