Предмет: Химия, автор: Анастасия0901

Сколько осадка выпало при сливании 200г 5% раствора хлорида железа с избытком раствора гидроксида натрия.

Ответы

Автор ответа: mrvladimir2
0
m(FeCl2) = 200 г*0,05=10 г
 10 г                        х г
FeCl2+2NaOH = Fe(OH)2+2NaCl
127 г                     90 г
х= 7,09 г
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