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9 August, 15:44

32 g of sulfur will react with 48 g of oxygen to produce 80 g of sulfur trioxide. If 32 g of sulfur and 100 g of oxygen are placed into a sealed container and allowed to react, what is the mass of the material in the container after the reaction is completed? (A) 100 g (B) 132 g (C) 80 g (D) 32 g

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  1. 9 August, 16:02
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    Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3 + 52 grams O2 = 132 grams (option B)

    Explanation:

    Step 1: Data given

    Mass of sulfur = 32.00 grams

    Mass of oxygen = 48.00 grams

    Molar mass of sulfur = 32.07 g/mol

    Molar mass of oxygen = 32 g/mol

    Molar mass of SO3 = 80.07 g/mol

    Step 2: The balanced equation

    2S + 3O2 → 2SO3

    Step 3: Calculate moles S

    Moles S = Mass S / molar mass S

    Moles S = 32.0 grams / 32.07 g/mol

    Moles S = 0.998 moles

    Step 4: Calculate moles O2

    Moles O2 = 100.0 grams / 32.0 g/mol

    Moles O2 = 3.125 moles

    Step 5: Calculate the limiting reactant

    For 2 moles S we need 3 moles O2 to produce 2 moles SO3

    S is the limiting reactant. It will completely be consumed (0.998 moles)

    O2 is in excess, there will be consumed 3/2 * 0.998 = 1.497 moles

    There will remain 3.125 - 1.497 = 1.628 moles O2

    This is 1.628 moles * 32 g/mol = 52.1 grams

    Step 6: Calculate moles SO3

    For 2 moles S we need 3 moles O2 to produce 2 moles SO3

    For 0.998 moles S there will react 0.998 moles SO3

    Step 6: Calculate mass SO3

    Mass SO3 = moles SO3 * molar mass SO3

    Mass SO3 = 0.998 moles * 80.07 g/mol

    Mass SO3 = 79.9 grams ≈ 80 grams

    There will be produced 80 grams of SO3

    Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3 + 52 grams O2 = 132 grams (option B)
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