Answer:
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 %
