\(n_{AlCl3}=\dfrac{26,7}{133,5}=0,2\left(mol\right)\)
\(n_{Al\left(OH\right)3}=\dfrac{7,8}{78}=0,1\left(mol\right)\)
Vì \(n_{Al\left(OH\right)3}< n_{AlCl3}\) nên có 2 gtr \(n_{KOH}\) thỏa mãn
PTHH : \(AlCl_3+3KOH-->Al\left(OH\right)_3+3KCl\) (1)
\(Al\left(OH\right)_3+KOH-->KAlO_2+2H_2O\) (2)
TH1 : KOH thiếu => Chỉ xảy ra pứ (1)
\(n_{KOH}=3n_{Al\left(OH\right)3}=0,3\left(mol\right)\)
=> \(C\%KOH=\dfrac{56.0,3}{400}.100\%=4,2\%=a\%\)
=> a = 4,2
TH2 : KOH dư => Xảy ra pứ (1) và (2)
Có : \(n_{KOH}=3n_{AlCl3}+n_{Al\left(OH\right)3tan}\)
\(=3n_{AlCl3}+\left(n_{Al\left(OH\right)3sinhra}-n_{Al\left(OH\right)3thuduoc}\right)\)
\(=3n_{AlCl3}+\left(n_{AlCl3}-n_{Al\left(OH\right)3thuduoc}\right)\)
\(=4n_{AlCl3}-n_{Al\left(OH\right)3thuduoc}\)
\(=4.0,2-0,1=0,7\left(mol\right)\)
=> \(C\%KOH=\dfrac{0,7.56}{400}.100\%=9,8\%=a\%\)
=> a = 9,8
