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12 October, 13:03

What mass of iron is needed to react with sulfer in order to produce 96 grand of Fe2S3 according to this equation: 2Fe + 3S = Fe2S3?

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Answers (2)
  1. 12 October, 13:17
    0
    The balanced equation for the reaction is as follows;

    2Fe + 3S - --> Fe₂S₃

    molar ratio of Fe to Fe₂S₃ is 2:1

    mass of Fe₂S₃ to be produced is - 96 g

    therefore number of moles of Fe₂S₃ to be produced is - 96 g / 208 g/mol

    number of Fe₂S₃ moles = 0.46 mol

    according to the molar ratio

    when 2 mol of Fe reacts with 3 mol of sulfur then 1 mol of Fe₂S₃ is produced

    that for 1 mol of Fe₂S₃ to be produced - 2 mol of Fe should react

    therefore for 0.46 mol of Fe₂S₃ to be produced - 2 x 0.46 = 0.92 mol of Fe is required

    mass of Fe required - 0.92 mol x 56 g/mol = 51.5 g

    mass of Fe required is - 51.5 g
  2. 12 October, 13:31
    0
    Answer: 51.69g

    Explanation:

    2Fe + 3S - > Fe2S3

    Molar Mass of Fe2S3 = (2x56) + (32x3) = 112 + 96 = 208g/mol

    Molar Mass of Fe = 56g/mol

    Mass conc. Of Fe = 2x56 = 112g

    From the equation,

    112g of Fe produced 208g of Fe2O3.

    Therefore, Xg of Fe will produce 96g of Fe2O3 i. e

    Xg of Fe = (112x96) / 208 = 51.69g
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