PTHH: M(OH)3 + 3HCl \(\rightarrow\) MCl3 + 3H2O (1)
Ta có: nHCl = \(\frac{90.7,3}{100.36,5}\)= 0,18 mol
Theo (1): nM(OH)3 = \(\frac{1}{3}\).nHCl
\(\Rightarrow\) \(\frac{6,42}{M+51}\) = \(\frac{1}{3}\).0,18
\(\Rightarrow\) \(\frac{6,42}{M+51}\) = 0,06
\(\Rightarrow\) 0,06M + 3,06=6,42
\(\Rightarrow\) M = 56
Vậy M là Fe.
PTHH: Fe(OH)3 + 3HCl \(\rightarrow\) FeCl3 + 3H2O (2)
Dung dịch A là FeCl3
Theo (2): nFeCl3 = \(\frac{1}{3}\).nHCl = \(\frac{1}{3}\).0,18= 0,06 mol
\(\Rightarrow\) mFeCl3 = 0,06.(56+35,5.3)=9,75g
mdd FeCl3 = 6,42+90= 96,42g
C%= \(\frac{9,75}{96,42}.100\) = 10,112%