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kitaykarina
@kitaykarina
July 2022
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В раствор массой 200г ,содержащий хлорид меди (II) массой 50г, поместили железный гвоздь массой 5г. Рассчитайте массовую долю хлорида меди (II) в растворе после окончания реакции
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vladislav9903
Fe+CuCl2=FeCl2+Cu N(Fe)= 5/56=0,09моль 0,09/1=х/1 х= 0,09
m(CuCl2) 0.09*134=13.4г 50-13,4=36,6
36,6/236,6=0,15 0,15*100=15%
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Answers & Comments
m(CuCl2) 0.09*134=13.4г 50-13,4=36,6
36,6/236,6=0,15 0,15*100=15%