bt hoá 9

T

thupham22011998

$Fe+CuSO_4-->FeSO_4+Cu$

Ta có: $n CuSO_4=0,1 mol$

Theo pthh, ta có: $n Fe=n Cu=n FeSO_4=0,1 mol$

$m$ d/d sau pư$=99,2g$

-->C%$FeSO_4=15,32$%
 
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