A fine of 50.0 mL of 0.0900 M CaCl2 reacts with excess sodium carbonate to give 0.366 g of calcium carbonate precipitate. What is the percen

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A fine of 50.0 mL of 0.0900 M CaCl2 reacts with excess sodium carbonate to give 0.366 g of calcium carbonate precipitate. What is the percent yield?

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MichaelMet 4 years 2021-08-31T08:21:57+00:00 1 Answers 24 views 0

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  1. Maris
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    2021-08-31T08:23:15+00:00
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    Answer:

    81.26% is the percent yield

    Explanation:

    Based on the reaction:

    CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

    Where 1 mole of CaCl₂ in excess of sodium carbonate produces 1 mole of calcium carbonate.

    To solve this question we must find the moles of CaCl2 added = Moles CaCO₃ produced (Theoretical yield). The percent yield is:

    Actual yield (0.366g) / Theoretical yield * 100

    Moles CaCl₂ = Moles CaCO₃:

    0.0500L * (0.0900moles / L) = 0.00450 moles of CaCO₃

    Theoretical mass -Molar mass CaCO₃ = 100.09g/mol-:

    0.00450 moles of CaCO₃ * (100.09g / mol) = 0.450g of CaCO₃

    Percent yield = 0.366g / 0.450g * 100

    81.26% is the percent yield

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