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11 January, 22:24

The percent yield of this reaction is consistently 92.0%. CH4 (g) + 4 S (g) → CS2 (g) + 2 H2S (g) How many grams of sulfur would be needed to obtain 72.57 g of CS2?

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  1. 11 January, 23:28
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    132.17 g

    Explanation:

    The reaction given, in the question is -

    CH₄ (g) + 4 S (g) - --> CS₂ (g) + 2H₂S (g)

    From the reaction, 4 mole of S is required for the production of 1 mole of CS₂.

    since,

    Moles of CS₂ = given mass of CS₂ / Molecular weight of CS₂

    Since,

    the Molecular weight of CS₂ = 76

    Given, mass of CS₂ = 72.57 g

    Moles of CS₂ = 72.57 / 76 = 0.95 mol

    Since,

    The yield is 92.0 %.

    Moles of S required = 4 * 0.95 mol / 0.92 = 4.13 moles

    Mass of S required = 4.13 * 32 = 132.17 g.
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