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What is the percent yield if 34.5 g of calcium hydroxide is produced when 37.6 g of calcium oxide reacts with 100 g of water in the following balanced equation?

Answer :

To find the percent yield, we first calculate the theoretical yield of calcium hydroxide to be 49.68 g. With an actual yield of 34.5 g, the percent yield is approximately 69.4%.

To determine the percent yield, we first need to calculate the theoretical yield, then use the actual yield to find the percent yield.

Step-by-Step Calculation:

  1. Write the balanced chemical equation:

    CaO + H₂O → Ca(OH)₂

  2. Calculate the molar masses:

    • Molar mass of CaO (Calcium oxide) = 40.08 g/mol (Ca) + 16.00 g/mol (O) = 56.08 g/mol
    • Molar mass of Ca(OH)₂ (Calcium hydroxide) = 40.08 g/mol (Ca) + 2(16.00 g/mol O + 1.01 g/mol H) = 74.10 g/mol

  3. Calculate moles of CaO used:

    Moles of CaO = 37.6 g / 56.08 g/mol = 0.6705 mol

  4. Calculate the theoretical yield of Ca(OH)₂:

    Since the reaction ratio is 1:1, moles of Ca(OH)₂ produced = moles of CaO used = 0.6705 mol

    Theoretical yield of Ca(OH)₂ = 0.6705 mol × 74.10 g/mol = 49.68 g

  5. Determine the percent yield using the actual yield:

    Percent yield = (actual yield / theoretical yield) × 100% = (34.5 g / 49.68 g) × 100% ≈ 69.4%

Therefore, the percent yield of calcium hydroxide in this reaction is about 69.4%.

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