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4 October, 19:29

In a laboratory experiment, when 35.0 g Mg reacted in excess O2, the percent yield of MgO was 90.0%. What was the actual yield of that experiment?

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  1. 4 October, 23:05
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    Actual yield = 54 g

    Explanation:

    Given dа ta:

    Mass of magnesium = 35.0 g

    Percent yield = 90.0%

    Actual yield = ?

    Solution:

    Chemical equation:

    2Mg + O₂ → 2MgO

    Number of moles of Mg:

    Number of moles = mass / molar mass

    Number of moles = 35.0 g / 24 g/mol

    Number of moles = 1.5 mol

    Now we will compare the moles of Mg and O.

    Mg : MgO

    2 : 2

    1.5 : 1.5 mol

    Mass of MgO: (theoretical yield)

    Mass = number of moles * molar mass

    Mass = 1.5 mol * 40 g/mol

    Mass = 60 g

    Percent yield = (actual yield / theoretical yield) * 100

    90.0 % = (actual yield / 60 g) * 100

    0.9 = actual yield / 60 g

    Actual yield = 0.9 * 60 g

    Actual yield = 54 g
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