Consider the chemical equation. CuCl2 + 2NaNO3 Cu(NO3)2 + 2NaCl What is the percent yield of NaCl if 31.0 g of CuCl2 reacts with excess NaNO3 to produce 21.2 g of NaCl? Use .

The percent yield of NaCl is 78.7 %

Explanation:

CuCl₂  +  2NaNO₃ →  Cu(NO₃)₂  + 2NaCl

If the NaNO₃ is determined to be in excess, the limiting reagent is the chloride. We convert the mass to moles:

31 g . 1mol / 134.45g = 0.230 moles

Ratio is 1:2, so we can make a rule of three to determine the theoretical yield

1 mol of copper (II) chloride reacts to produce 2 moles of sodium chloride

Then, 0.230 moles of CuCl₂ will react to produce (0.230 .2) /1 ) = 0.461 moles of NaCl → we convert the moles to mass → 0.461 mol . 58.45 g / 1mol = 26.9 g

To find percent yield we do → (Yield produced / Theoretical yield) . 100

(21.2 g / 26.9 g) . 100 = 78.7 %

answer; i believe the correct answer is (solid gold);

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error