\(n_{AlCl_3}=0,2.0,1=0,02\left(mol\right)\\ n_{Al\left(OH\right)_3}=\dfrac{0,78}{78}=0,01\left(mol\right)\)
TH1: AlCl3 dư, kết tủa không bị tan
PTHH:
\(2K+2H_2O\rightarrow2KOH+H_2\)
\(3KOH+AlCl_3\rightarrow Al\left(OH\right)_3\downarrow+3KCl\)
0,03<-----0,01<------0,01
nAlCl3 (pư) = 0,01 (mol) < 0,02 = nAlCl3 (bđ) (TM)
BTNT K: nK = nKOH = 0,03 (mol)
=> a = 0,03.39 = 1,17 (g)
TH2: KOH dư, kết tủa bị tan một phần
PTHH:
\(2K+2H_2O\rightarrow2KOH+H_2\)
\(3KOH+AlCl_3\rightarrow Al\left(OH\right)_3\downarrow+3KCl\)
0,06<----0,02------->0,02
\(KOH+Al\left(OH\right)_3\rightarrow KAlO_2+2H_2O\)
0,01<----0,01
BTNT K: nK = nKOH = 0,01 + 0,06 = 0,07 (mol)
=> a = 0,07.39 = 2,73 (g)
\(2K+2H_2O\rightarrow2KOH+H_2\\ 3KOH+AlCl_3\rightarrow Al\left(OH\right)_3\downarrow+3KCl\\ m_{Al\left(OH\right)_3}=m_{kt}=0,78\left(g\right)\\ \Rightarrow n_{Al\left(OH\right)_3}=\dfrac{0,78}{78}=0,01\left(mol\right)\\ n_K=n_{KOH}=3.n_{Al\left(OH\right)_3}=3.0,01=0,03\left(mol\right)\\ m_K=0,03.39=1,17\left(g\right)\\ \Rightarrow a=1,17\left(g\right)\)
\(V_{AlCl_3}=200ml=0,2\left(l\right)\)
=> \(n_{AlCl_3}=CM.V=0,1.0,2=0,02\left(mol\right)\)
\(3K+AlCl_3\rightarrow Al+3KCl\)
kết tủa là Al
từ pt suy ra : \(n_{KCl}=3.0,02=0,06\left(mol\right)\)
=> \(m_{KCl}=0,06.74,5=4,47\left(g\right)\)
\(m_{AlCl_3}=0,02.133,5=2,67\left(g\right)\)
Áp dụng đl BTKL:
\(m_K+m_{AlCl_3}=m_{Al}+m_{KCl}\)
\(m_K+2,67=0,78+4,47\)
=> \(a=m_K=\left(0,78+4,47\right)-2,67=2,58\left(g\right)\)
