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24 December, 12:58

A student mixes a solution containing 10.0 g bacl2 (m = 208.2) with a solution containing 10.0 g na2so4 (m = 142.1) and obtains 12.0 g baso4 (m = 233.2). what is the percent yield of this reaction?

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  1. 24 December, 14:29
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    The balanced equation for the above reaction is as follows;

    Na₂SO₄ + BaCl₂ - - > BaSO₄ + 2NaCl

    Na₂SO₄ reacts with BaCl₂ in the molar ratio 1:1

    Number of Na₂SO₄ moles - 10.0 g / 142.1 g/mol = 0.0704 mol

    Number of BaCl₂ moles - 10.0 g / 208.2 g/mol = 0.0480 mol

    this means that 0.0480 mol of each reactant is used up, BaCl₂ is the limiting reactant and Na₂SO₄ has been provided in excess.

    stoichiometry of BaCl₂ to BaSO₄ is 1:1

    number of BaSO₄ moles formed - 0.0480 mol

    Mass of BaSO₄ - 0.0480 mol x 233.2 g/mol = 11.2 g

    theoretical yield is 11.2 g but the actual yield is 12.0 g

    the actual product maybe more than the theoretical yield of the product as the measured mass of the actual yield might contain impurities.

    percent yield - 12.0 g / 11.2 g x 100% = 107%

    this is due to impurities present in the product or product could be wet.
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