Heating copper(II) sulfate pentahydrate will evaporate the water from the crystals, leaving anhydrous copper sulfate, a white powder. Anhydrous means “without water”. What mass of anhydrous copper(II) sulfate would be produced by heating 500.0 g of CuSO 4 · 5H 2 O?
The answer is
% CuSO 4 = 63.92%
% H 2 O = 36.08%
Mass of anhydrous copper(ii) sulfate = 319.6g
Im not sure how to solve this though
2 answers
500 g CuSO4.5H2O x 0.6392 g CuSO4/1 g CuSO4.5H2O) = ? g CuSO4
CuSO4 = 63.5 + 32 + 4*16 = 159.5 grams/mol
5 H2O = 5 ( 2+16) = 90 grams / mol
so CuSO4 5H2O = 249.5 g/mol
so
CuSO4/total = 0.639 or 63.9 %
H2O/total = 0.361 or 36.1 %
5 H2O = 5 ( 2+16) = 90 grams / mol
so CuSO4 5H2O = 249.5 g/mol
so
CuSO4/total = 0.639 or 63.9 %
H2O/total = 0.361 or 36.1 %
4 answers