If 0.673g of Cu is reacted with excess nitric acid to form 0.990g of copper nitrate, what is the percent yield of copper nitrate in this reaction, from the stoichiometry, it is known tgat 1.000g Cu should form 2.952g of copper nitrate?
2006-09-27 17:03:42 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
If 1.000 g should give 2.952 g, then 0.673 g would give:
0.673 * 2.952 = 1.987 g this is called the theoretical yield.
What you got was 0.990 g this is the actual yield.
The percent yield is actual yield / theoretical yield * 100
= 0.990 g / 1.987 g * 100 = percent yield
I am sure you can figure it out from here.
2006-09-27 17:12:56 · answer #1 · answered by Richard 7 · 68⤊ 0⤋
The theoretical weight % yield is 2.952*100 as stated, but the actual obtained is 0.99/0.673*100= 147 which is very poor (147/295= 0.498) or 50% than theoretical.
2006-09-28 00:24:34 · answer #2 · answered by Fred 2 · 0⤊ 0⤋
This is simply a proportionality problem, and requires nothing more to solve than a four-banger with a working battery.
2006-09-28 00:06:36 · answer #3 · answered by Anonymous · 0⤊ 0⤋
% yield = actual amount/theorized amount X100
2006-09-28 00:06:10 · answer #4 · answered by Greg G 5 · 0⤊ 0⤋