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6 August, 04:33

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?

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  1. 6 August, 04:36
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    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.
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