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If 11.3 grams of sodium chloride are formed in the reaction, what is the percent yield of this reaction?1CuCl2 + 2NaNO3 → 1Cu(NO3)2 + 2NaCl15 grams of copper (II) chloride react with 20 grams of sodium nitrate

User Ananda Pramono
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1 Answer

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Step 1

The reaction:

1CuCl2 + 2NaNO3 → 1Cu(NO3)2 + 2NaCl (balanced)

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Step 2

Information provided: the actual yield = 11.3 g NaCl (sodium chloride)

Mass of CuCl2 = 15 g and mass of NaNO3 = 20 g

Information needed: the limiting reactant, and the molar masses of:

CuCl2) 134.4 g/mol

NaNO3) 85.0 g/mol

NaCl) 58.4 g/mol

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Step 3

The limiting reactant and the excess: Procedure

134.4 g CuCl2 -------------- 2 x 85.0 g NaNO3

15 g CuCl2 ------------- X = 18.9 g NaNO3

For 15 g CuCl2, 18.9 g NaNO3 is needed, therefore the excess is NaNO3, and the limiting reactant is the CuCl2.

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Step 4

The theoretical yield: mass of NaCl formed from the stoichiometry

134.4 g CuCl2 ------------ 2 x 58.4 g NaCl

15 g CuCl2 ------------ X = 13.0 g = theoretical yield

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Step 5

The % yield = actual yield/theoretical yield x 100 = 11.3 g/13.0 g x 100 = 86.9 %

Answer: % yield = 86.9 %

User Kkgarg
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