\(2Al+6HCl-->2AlCl_3+3H_2\left(1\right)\)

0,04___ 0,12_________0,04______ 0,06

\(Al_2O_3+6HCl-->2AlCl_3+3H_2O\left(2\right)\)

0,02_____0,13________ 0,04

\(n_{H_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

\(n_{HCl}=0,25.1=0,25\left(mol\right)\)

=> \(n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.0,06=0,04\)

\(n_{HCl\left(1\right)}=2n_{H_2}=2.0,06=0,12\left(mol\right)\)

=> \(n_{HCl\left(2\right)}=0,25-0,12=0,13\left(mol\right)\)

=> \(n_{Al_2O_3}=\dfrac{1}{6}n_{HCl\left(2\right)}=\dfrac{1}{6}.0,13=0,02\left(mol\right)\)

=> \(m_{Al}=0,04.27=1,08\left(gam\right)\)

=> %\(m_{Al}=\dfrac{1,08}{2,1}.100=51,43\%\)

=>%\(m_{Al_2O_3}=100-51,43=48,57\%\)

=> Σ \(n_{AlCl_3}=0,0+0,04=0,08\left(mol\right)\)

=> \(C_M\left(AlCl_3\right)=\dfrac{0,08}{0,25}=0,32\left(M\right)\)

\(AlCl_3+3NaOH-->Al\left(OH\right)_3+3NaCl\left(3\right)\)

0,08_______0,24________0,08

\(n_{Al\left(OH\right)_3}=\dfrac{3,9}{78}=0,05\left(mol\right)\)

=> \(n_{Al\left(OH\right)_3}=n_{AlCl_3}=0,08>0,05\left(mol\right)\)

=> Al(OH)₃ bị tan một phần

=> \(n_{Al\left(OH\right)_3}bịtan=0,08-0,05=0,03\left(mol\right)\)

\(NaOH+Al\left(OH\right)_3-->NaAlO_2+2H_2O\left(4\right)\)

0,03______ 0,03

=> Σ \(n_{NaOH}=0,24+0,03=0,27\left(mol\right)\)

=> \(V_{NaOH}=\dfrac{0,27}{1,6}=0,16875\left(l\right)\)

a)R + H₂SO₄ => RSO₄ + H₂ ⁽¹⁾

Đổi 672 ml =0,672 l

n_H2 = 0,03 (mol)

(1) => n_R = n_H2 = 0,03 (mol)

=> M_R = \(\frac{1,95}{0,03}=65\) (g/mol)

=> R : Zn

b) H₂SO₄ + 2NaOH=>Na₂SO₄ + 2H₂O ⁽²⁾

Đổi 250 ml = 0,25 l

n_NaOH = 0,25.0,2 = 0,05 (mol)

(2) => n_{H2SO4 dư =} \(\frac{1}{2}.n_{NaOH}=\frac{0,05}{2}=0,025\left(mol\right)\)

(1) => n_H2SO4(1) = n_H2 = 0,03 (mol)

=> n_{H2SO4 đã dùng} = 0,03+0,025 = 0,055(mol)

=> V = 0,055:0,2 = 0,275 (l)

c) Zn + 2KOH => K₂ZnO₂ + H₂ (3)

(3) => n_KOH = 2.n_Zn = 2.0,03 = 0,06 (mol)

m_{dd KOH} = \(\frac{0,06.56}{33,6\%}=10\left(g\right)\)

hh X????

Đầu bài ở trên là "Hòa tan hoàn toàn 11 g hỗn hợp" nhé!!