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1 July, 17:47

A chemist reacted 17.25 grams of sodium metal with an excess amount of chlorine gas. The chemical reaction that occurred is shown.

Na + Cl2 → NaCl

If the percentage yield of the reaction is 88%, what is the actual yield? Show your work, including the use of stoichiometric calculations and conversion factors.

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  1. 1 July, 17:50
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    Actual yield for the reaction is 38.5 g of NaCl

    Explanation:

    The balanced equation is this:

    2 Na (s) + Cl₂ (g) → 2NaCl (s)

    Ratio is 2:2. Let's determine the moles of sodium metal that has reacted.

    17.25 g. 1 mol / 23 gl = 0.75 moles

    0.75 moles of Na were used to produced 0.75 moles of NaCl, according to the equation.

    Let's determine the mass of NaCl, that were produced.

    0.75 mol. 58.45 g / 1 mol = 43.8 g

    43.8 will be the theoretical yield (100%). Let's determine the 88%.

    (Yield reaction / Theoretical yield). 100 = 88

    Yield reaction / 43.8 g = 88 / 100

    Yield reaction = 0.88. 43.8 g → 38.5 g
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