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 %

beta decay is the answer

explanation:

0.07 g/cm³

1.37 g/cm³