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27 August, 19:43

Consider the chemical equation. CuCl2 + 2NaNO3 Cu (NO3) 2 + 2NaCl What is the percent yield of NaCl if 31.0 g of CuCl2 reacts with excess NaNO3 to produce 21.2 g of NaCl? Use.

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  1. 27 August, 21:05
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    The percent yield of NaCl is 78.7 %

    Explanation:

    CuCl₂ + 2NaNO₃ → Cu (NO₃) ₂ + 2NaCl

    If the NaNO₃ is determined to be in excess, the limiting reagent is the chloride. We convert the mass to moles:

    31 g. 1mol / 134.45g = 0.230 moles

    Ratio is 1:2, so we can make a rule of three to determine the theoretical yield

    1 mol of copper (II) chloride reacts to produce 2 moles of sodium chloride

    Then, 0.230 moles of CuCl₂ will react to produce (0.230.2) / 1) = 0.461 moles of NaCl → we convert the moles to mass → 0.461 mol. 58.45 g / 1mol = 26.9 g

    To find percent yield we do → (Yield produced / Theoretical yield). 100

    (21.2 g / 26.9 g). 100 = 78.7 %
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