Emily P.

asked • 05/20/21

Chemistry question

What is the greatest amount of sodium chloride in moles that can be made with 13 grams copper II chloride and 15 grams sodium nitrate? Which reactant is limiting? How many moles of the excess reactant will be left after the reaction?


CuCl2  + 2 NaNO3  --->Cu(NO3)2 + 2 NaCI

Alex O.

Here is my understanding of it for the part that asks what is the limiting reactant. You have two options, either CuCl2 and NaNO3. Here is how to find out. Fill out the numbers below with the molar masses: 13gCuCl2*(1mol/molar mass CuCl2)*(2mol NaCl/1molCuCl2) = X 15gNaNo2*(1mol/molar mass NaNO3)*(2mol NaCl/1molNaNO3) = Y If X > Y, then (NaNO2) is the limiting reactant. If Y> X, then (CuCl2) is the limiting reactant. Let me know if you have any questions on this part.
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05/21/21

1 Expert Answer

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Sidney P. answered • 05/22/21

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(13g CuCl2) * (1 mole CuCl2 / 134.4g CuCl2) * (2 mole NaCl / 1 mole CuCl2) = 0.193 mole NaCl.

(15g NaNO3) * (1 mole NaNO3 /85.0g NaNO3) * (2 mole NaCl / 2 mole NaNO3) = 0.176 mole NaCl.

Sodium nitrate is limiting, producing 0.18 mole NaCl, with 0.017 mole excess of copper chloride (this result has only 1 sig fig, 0.02 mole, but the answer would be wrong if you rounded the individual mole numbers before taking the difference).

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