Sulfur trioxide, ammonia gas, and water will react to form ammonium sulfate. Determine the percent yield when 3.78 g of ammonium sulfa

Question

Sulfur trioxide, ammonia gas, and water will react to form ammonium sulfate. Determine the
percent yield when 3.78 g of ammonium sulfate forms from the reaction of 2.51 g of sulfur
trioxide, 1.55 g of ammonia, and 0.804 g of water.

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Kim Cúc 5 years 2021-07-19T13:48:47+00:00 1 Answers 39 views 0

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  1. Sapo1
    0
    2021-07-19T13:49:51+00:00
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    Answer:

    Explanation:

    Sulfur trioxide, ammonia gas, and water will react to form ammonium sulfate. Determine the percent yield when 3.78 g of ammonium sulfate forms from the reaction of 2.51 g of sulfur trioxide, 1.55 g of ammonia, and 0.804 g of water. Can you please show step by step on how you got the answer thanks!

    –

    First, the balanced equation:

    1 SO3 + 2 NH3 + 1 H2O –> 1 (NH4)2SO4

    Find the limiting reactant using molar masses and mole ratios

    2.51 g SO3 (1 mol SO3 / 80.06 g SO3) (1 mol (NH4)2SO4 / 1 mol SO3) = 0.03135 mol (NH4)2SO4

    1.55 g NH3 (1 mol NH3 / 17.04 g NH3) (1 mol (NH4)2SO4 / 2 mol NH3) = 0.04548 mol (NH4)2SO4

    0.804 g H2O (1 mol H2O / 18.02 g H2O) (1 mol (NH4)2SO4 / 1 mol H2O) = 0.04462 mol (NH4)2SO4

    Therefore, the limiting reactant is sulfur trioxide.

    Now find the theoretical yield of (NH4)2SO4.

    2.51 g SO3 (1 mol SO3 / 80.06 g SO3) (1 mol (NH4)2SO4 / 1 mol SO3) (132.16 g (NH4)2SO3 / 1 mol (NH4)2SO4) = 4.14 g (NH4)2SO4

    Actual yield / Theoretical yield x 100% = Percent yield.

    Therefore, percent yield = 91.3%

    hoppe it helps

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