Carbon disulfide is prepared by heating sulfur and charcoal. the chemical equation is s 2 ( g ) + c ( s ) − ⇀ ↽ − cs 2 ( g ) k c = 9.40 at 900 k how many grams of cs 2 ( g ) can be prepared by heating 14.2 mol s 2 ( g ) with excess carbon in a 6.30 l reaction vessel held at 900 k until equilibrium is attained?

974.6 grams of CS2 can be prepared.

Explanation:

Step 1: 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

Step 2: The balanced equation:

S2(g)+C(s) ↔ CS2(g)

Step 3: 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

Step 4: 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

Step 5: Calculate moles of CS2

Moles CS2 = molarity CS2 * volume

Moles CS2 = 2.034 M * 6.30 L

Moles CS2 = 12.8 moles

Step 6: 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.

am i too late?

explanation:

he would use the colour black