Question

If 18.5 grams of CuCl2 react with 22.8 grams of NaNO3, what mass of NaCl can be formed?

CuCl2 + NaNO3 → Cu(NO3)2 + NaCl

12.4 g NaCl
15.7 g NaCl
16.2 g NaCl
28.4 g NaCl

Answer 1

From above calculations, gives the limited amount of NaCl. So, 15.7 g of NaCl would form from given amounts of reactants.

Step-by-step explanation:

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