Question

Hòa tan 3,72g hh (AL, AL2O3) TRONG 416g dd HCl (D=1.04 g/ml) vừa đủ thu được 1,848 lít khí (27.3 độ C: 2atm)

a)Tính % khối lượng mỗi chất có trong hh

b)Tính CM dd HCl và C% dung dịch muối thu được

Answer 1

a)2Al+6HCl-->2AlCl3+3H2

0,055--------------------0,0825 mol

nH2=1,848\22,4=0,0825 mol

=>mAl=0,055.27=1,485 g

=>%mAl=1,485\3,75.100=39,6 %

=>%mAl2O3=100-39,6=60,4 %

Answer 2

a) \(2Al+6HCl-->2AlCl3+3H2\)

\(Al2O3+6HCl-->2AlCl3+3H2O\)

\(n_{H2}=\frac{1,848}{22,4}=0,0825\left(mol\right)\)

\(n_{Al}=\frac{2}{3}n_{H2}=0,055\left(mol\right)\)

\(\%m_{Al}=\frac{0,055.27}{3,72}.100\%=40\%\)

\(\%m_{Al2O3}=100-40=60\%\)

b) \(n_{HCl\left(1\right)}=2n_{H2}=0,165\left(mol\right)\)

\(m_{Al2O3}=3,72-0,055.27=2,235\left(mol\right)\)

\(n_{Al2O3}=\frac{2,235}{102}=0,022\left(mol\right)\)

\(n_{HCl}=6n_{Al2O3}=0,132\left(mol\right)\)

\(\sum n_{HCl}=0,132+0,165=0,297\left(mol\right)\)

\(V_{HCl}=\frac{416}{1,04}=400ml=0,4\left(l\right)\)

\(C_{M\left(HCl\right)}=\frac{0,297}{0,4}=0,7425\left(M\right)\)

\(m_{dd}=m_{hh}+m_{ddHCl}-m_{H2}=3,72+416-0,165=419,555\left(g\right)\)

\(m_{AlCl3\left(1\right)}=0,055.133,5=7,3425\left(g\right)\)

\(m_{AlCl3\left(2\right)}=0,044.133,5=5,874\left(g\right)\)

\(\sum m_{AlCl3}=5,874+7,3425=13,2165\left(g\right)\)

\(C\%_{AlCl3}=\frac{13,2165}{419,555}.100\%=3,15\%\)