When 15 grams of copper (II) chloride (CuCl2) reacts with 20 grams of sodium nitrate (NaNO3), 11.3 grams of sodium chloride (NaCl) is obtained. What is the percent yield of this reaction?

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

Proportionally, for every one mole of sodium nitrate, you should yield one mole of sodium chloride.

Sodium nitrate's molar mass is 85.0 grams/mole:

20 grams / (85.0 grams/mole) = 0.24 moles of sodium nitrate

The theoretical yield of sodium chloride would also be 0.24 moles

The molar mass of sodium chloride is 58.5 grams/mole:

11.3 grams / (58.5 grams/mole) = 0.19 moles

Percent yield = ((actual yield) / (theoretical yield)) * 100%

Percent yield = (0.19 moles/0.24 moles) * 100% = 79% yield