a)\(2Al+6HCl\xrightarrow[]{}2AlCl_3+3H_2\\ Fe+2HCl\xrightarrow[]{}FeCl_2+H_2\)
Gọi x,y là số mol Al và Fe
\(n_{H_2}=1,5n_{Al}+n_{Fe}=1,5x+y=\dfrac{8,96}{22,4}=0,4\left(mol\right)\) (1)
Ta có: \(m_{Al+Fe}=27x+56y=14,2-3,2=11\left(g\right)\) (2)
Từ (1) và (2) giải hệ ⇒ \(x=0,2\\ y=0,1\)
\(\%^m_{Al}=\dfrac{0,2.27}{14,2}.100=38,03\%\)
\(\%^m_{Fe}=\dfrac{0,1.56}{14,2}.100=39,44\%\)
\(\%^m_{Cu}=100-38,03-39,44=22,53\%\)
b) 14,2g hh X có 5,4g Al, 5,6g Fe và 3,2g Cu
⇒ 7,1g hh X có 2,7g Al, 2,8g Fe và 1,6g Cu
\(2Al+6H_2SO_{4_{ }đặc}\xrightarrow[]{t^o}Al_2\left(SO_4\right)_3+3SO_2+6H_2O\\ 2Fe+6H_2SO_{4_{ }đặc}\xrightarrow[]{t^o}Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\\ Cu+2H_2SO_{4_{ }đặc}\xrightarrow[]{t^o}CuSO_4+SO_2+2H_2O\)
\(n_{SO_2}=1,5n_{Al}+1,5n_{Fe}+n_{Cu}=1,5.\dfrac{2,7}{27}+1,5.\dfrac{2,8}{56}+\dfrac{1,6}{64}=0,25\left(mol\right)\)
\(V_{SO_2\left(đktc\right)}=0,25.22.4=5,6\left(l\right)\)
c) \(Cu+4HNO_{3_{ }đặc}\xrightarrow[]{}Cu\left(NO_3\right)_2+2NO_2+2H_2O\)
\(n_{NO_2}=2n_{Cu}=2.0,025=0,05\left(mol\right)\)
\(V_{NO_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\)
d)dd A gồm \(FeCl_2\), \(AlCl_3\) và HCl dư
\(AgNO_3+HCl\xrightarrow[]{}AgCl+HNO_3\) (*)
\(AlCl_3+3AgNO_3\xrightarrow[]{}Al\left(NO_3\right)_3+3AgCl\) (**)
\(FeCl_2+2AgNO_3\xrightarrow[]{}Fe\left(NO_3\right)_2+2AgCl\) (***)
\(Fe\left(NO_3\right)_2+AgNO_3\xrightarrow[]{}Fe\left(NO_3\right)_3+Ag\)
\(m_{FeCl_2}+m_{AlCl_3}=0,1.127+0,2.133,5=39,4\left(g\right)\)
Theo ĐLBTKL: \(m_{Al+Fe}+m_{HCl}=m_{FeCl_2+AlCl_3}+m_{H_2}\)
⇒ \(m_{HCl}=29,2\left(g\right)\)
\(m_{HCl_{ }dư}=\dfrac{29,2.100}{90}-29,2=3,2\left(g\right)\)
\(n_{AgCl\left(\cdot\right)}=n_{HCl}=\dfrac{3,2}{36,5}=0,09\left(mol\right)\)
\(n_{AgCl\left(\cdot\cdot\right)}=3n_{AlCl_3}=3.0,2=0,6\left(mol\right)\)
\(n_{AgCl\left(\cdot\cdot\cdot\right)}=2n_{FeCl_2}=2n_{Fe\left(NO_3\right)_2}=2.0,1=0,2\left(mol\right)\)
\(n_{Ag}=n_{Fe\left(NO_3\right)_2}=0,2\left(mol\right)\)
\(m\downarrow=\left(0,9+0,6+0,2\right).143,5+0,2.108=265,55\left(g\right)\)
