cho 16,25g FeCl3 td với dd Na2S dư

H

hung_ils

nFeCl3=0,1
2FeCl3+Na2S\Rightarrow2FeCl2+2NaCl+S
0,1---------------0,1-------0,05
FeCl2+Na2S\Rightarrow2NaCl+FeS
0,1-------------------0,1
\Rightarrowm(kết tủa)=mS+mFeS=0,05*32+0,1*(56+32)=10,4\RightarrowA
 
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