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olgashar1976
@olgashar1976
July 2022
1
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сколько граммов FeCi3* 7H2O можно получить из 1 л 12%-ного раствора безводной соли FeSo4, плотность которой 1,122 г/мл ?
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tomochkasinits
M(FeSO4)р-ра=1000*1,122=1122г; m(FeSO4)в р-ре=1122*0,12=448,8;
n(FeSO4)=448,8/152=3моль; n(Fe)=3моль; m(Fe)=3*56=168г;
W(Fe)в FeCl3*7H2O=56/288,5=0,194;
m(FeCl3*7H2O)=168/0,194=866гр.
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Answers & Comments
n(FeSO4)=448,8/152=3моль; n(Fe)=3моль; m(Fe)=3*56=168г;
W(Fe)в FeCl3*7H2O=56/288,5=0,194;
m(FeCl3*7H2O)=168/0,194=866гр.