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30 June, 08:03

A student completes a lab and produces 1.3 moles of hydrogen gas by adding magnesium to sulfuric acid. If the student started with 22.8 grams of magnesium, what was the student's percent yield?

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  1. 30 June, 08:05
    0
    72 %

    Step-by-step explanation:

    We know we will need a balanced equation with masses and molar masses, so let's gather all the information in one place.

    M_r: 24.30

    Mg + H₂SO₄ ⟶ MgSO₄ + H₂

    m/g: 22.8

    n/mol: 1.3

    Calculations:

    (a) Moles of Mg

    n = 22.8 g Mg * (1 mol Mg/24.30 g Mg)

    = 0.9383 mol Mg

    (b) Moles of H₂

    The molar ratio is (1 mol H₂/1 mol Mg).

    n = 0.9383 mol Mg * (1 mol H₂/1 mol Mg)

    = 0.9383 mol H₂

    (c) Percent yield

    % yield = actual yield/theoretical yield * 100 %

    = 0.9383 mol/1.3 mol * 100 %

    = 72 %
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