CuCl2 + 2AgNO3 --> Cu(NO3)2 + 2AgCl
to three places:
moles AgNO3 = grams/molar mass = 6.10/170 = 0.0359
moles AgCl formed = 2*0.0359/2 = 0.0359
g AgCl formed = moles AgCl x molar mass AgCl = 0.0359*143 = 5.13 g = theoretical yield = TY
Actual yield from problem = AY = 4.77 g.
% yield = (AY/TY)*100 = ?
For the following reaction, 6.10 grams of silver nitrate are mixed with excess copper(II) chloride. The reaction yields 4.77 grams of silver chloride
What is the theoretical yield of silver chloride? What is the percent yield of silver chloride?
1 answer
1 answer