n FeCl3.6H2O=0,1 =nFeCl3
nNaOH=\(\frac{100.20}{100.40}=0,5mol\)
.......................FeCl3 + 3 NaOH → Fe(OH)3 + 3NaCl
trc pứ .............0,1 ........0,5......... ........ 0.................0
pứ .................. 0,1.........0,3...................0,1...............0,3
sau pứ..............0............0,2.................0,1................0,3
m kết tủa=0,1.(56+17.3)=10,7 g
b. m dd sau pứ= 27,05 + 100=127,05 g
C%m NaOH dư=\(\frac{0,2.40}{127,05}.100\%=6,3\%\)
C%mNaCl=\(\frac{0,3.58,5}{127,05}.100\%=13,8\%\)