Question

What is the limiting reagent for the reaction, how much of the excess reagent is left over in this reaction? CuCl2+2NaNO3=Cu(NO3)2+2NaCl15 grams of copper chloride with 20 grams of sodium nitrate 11.3 grams of sodium chloride is formed

Answer 1

Step 1

The reaction must be written and balanced:

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

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Step 2

The limiting reagent

Information needed:

The molar masses of CuCl2 (134,4 g/mol) and NaNO3 (84,9 g/mol)

By stoichiometry:

1 mole CuCl2 = 134.4 g

1 mole NaNO3 = 84.9 g

134.4 g CuCl2 ---------- 2 x 84.9 g NaNO3

15 g CuCl2 ---------- X = 18.95 g NaNO3

For 15 g CuCl2, 18.95 g of NaNO3 is needed, but there is 20 g. Therefore, The limiting reagent is CuCl2 and the excess is NaNO3

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Step 3

Amount of the excess reagent leftover:

20 g NaNO3 - 18.95 g NaNO3 = 1.05 g leftover

Answer:

The limiting reagent = CuCl2

The amount of excess leftover = 1.05 g NaNO3