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16 January, 00:31

In an experiment, 22.00 mL of a 0.250 M calcium chloride are combined with 35.00 mL of a 0.200 M sodium phosphate solution to produce a white precipitate. After the precipitate is collected and dried, it is found to have a mass of 0.973 g. What is the percent yield of this reaction?

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  1. 16 January, 03:15
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    Percent yield = (Actual yield / Theoritical yield) x 100%

    Actual yield is given as 0.973 g

    Calculation of theoritical yield;

    3CaCl₂ (aq) + 2Na₃PO₄ (aq) → Ca₃ (PO₄) ₂ (s) + 6NaCl (aq)

    Stoi. ratio 3 : 2 : 1

    Initial moles 5.5 x 10⁻³ 7 x 10⁻³

    Reacted 5.5 x 10⁻³ (5.5 x 10⁻³ x 2) / 3

    Final moles - 3.33 x 10⁻³ (5.5 x 10⁻³) / 3

    to calculate the moles following equation was used.

    moles = volume (L) x Molarity (mol L⁻¹)

    The molar mass of Ca₃ (PO₄) ₂ (s) = 310 g mol⁻¹

    mass of Ca₃ (PO₄) ₂ (s) = moles x molar mass

    = (5.5 x 10⁻³) / 3 mol x 310 g mol⁻¹

    = 0.568 g

    percent yield = (0.973 g / 0.568 g) x 100%

    = 171.30 %

    Since the percent yield is over that 100, there is an error in actual yield.

    the error can be in measuring instrument or the product is not dried well.
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