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2 December, 11:03

When copper is heated with an excess of sulfur, copper (I) sulfide is formed. In a given experiment, 0.0970 moles of copper was heated with excess sulfur to yield 1.76 g copper (I) sulfide. What is the percent yield?

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  1. 2 December, 11:16
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    Percent yield = 22.8 %

    Explanation:

    Step 1: Data given

    Numbers of moles copper = 0.0970 moles

    Mass of copper (I) sulfide = 1.76 grams

    Step 2: The balanced equation

    2Cu + S ⇒ Cu2S

    Step 3: Calculate moles of Cu2S

    For 2 moles Cu we need 1 mol S to produce 1 mol Cu2S

    For 0.0970 moles Cu we'll hace 0.0970 / 2 = 0.0485 moles

    Step 4: Calculate mass of Cu2S

    Mass Cu2s = moles Cu2S * molar mass Cu2S

    Mass Cu2S = 0.0485 moles * 159.16 g/mol

    Mass Cu2S = 7.72 grams

    Step 5: Calculate percent yield

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

    Percent yield = (1.76 grams / 7.72 grams) * 100%

    Percent yield = 22.8 %
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