what is the percent yield of CuS for the following reaction given that you start with 15.5 grams of Na2S and 12.1 grams CuSO4? The actually amount of CuS produced was 3.05 grams Reaction: Na2S + CuSO4>>>Na2SO4 + CuS
2007-04-27 20:16:25 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
1. work out the moles of Na2S
2. work out the moles of CuSO4
3. work out the moles of CuS
4. select which is smaller of 1 and 2, divide 3 by this value and multiply by 100.
2007-04-27 20:35:19 · answer #1 · answered by Gervald F 7 · 0⤊ 1⤋
Na2S + CuSO4 ----->Na2SO4 + CuS
Percentage yield = ( actual yield of product / theoretical yield of product) x 100
% yield = 3.05/15.5 * 100
= 19.7%
% yield = 3.05/12.1 * 100
=25.2%
2007-04-28 03:52:43 · answer #2 · answered by Tubby 5 · 0⤊ 0⤋
13.2 %
2007-04-28 04:02:45 · answer #3 · answered by ag_iitkgp 7 · 0⤊ 0⤋