2Al+6HCl→2AlCl3+3H2

nH2=5,376/22,4=0,24mol

nAl=0,24.2/3=0,16mol

nHCl=0,16.3=0,48mol

mAl=0,16.27=4,32g

mAlCl3=0,16.133,5=21,36g

mdd sau pứ=4,32+200-0,24.2=203,84g

C%AlCl3=21,36/203,84.100%=10,48%

mHCl dư=(0,48.15/100).36,5=2,628g

C%HCl dư=2,628/203,84.100%=1,29%

$\text{Câu a:}$

$\text{$n_{H{2}}$ = 5.376/22.4 = 0.24 mol}$

$\textit{2Al + 6HCl → $2AlCl_{3}$ + $3H_{2}↑$}$

$\text{$n_{Al}$ = 2/3$n_{H{2}}$ = 2/3*0.24 = 0.16 mol ⇒ m = $m_{Al}$ = 0.16*27 = 4.32 g}$

$\text{Câu b:}$

$\text{$n_{HCl}p. ứ.$ = 6/3$n_{H{2}}$ = 6/3*0.24 = 0.48 mol}$

$\text{C% $dd_{HCl}$ ban đầu = [(0.48*36.5)/200]*100% + 15% = 23.76 %}$

$\text{⇒ $m_{HCl}$ = 200*23.76/100 = 47.52 g ⇒ $n_{HCl}$ = 47.52/36.5 = 1.3 mol}$

$\text{⇒ $n_{HCl}$dư = $n_{HCl}$ - $n_{HCl}p. ứ.$ = 1.3 - 0.48 = 0.82 mol}$

$\text{$n_{AlCl{3}}$ = 2/3$n_{H{2}}$ = 2/3*0.24 =0.16 mol}$

$\text{$m_{dd}$ = $m_{Al}$ + $m_{dd{HCl}}$ - $m_{H{2}}$ = 4.32 + 200 - 0.48 = 203.84 g}$

$\text{C% $dd_{HCl}$dư = (0.82*36.5)/203.84*100% = 14.68%}$

$\text{C% $dd_{AlCl{3}}$ = (0.16*133.5)/203.84*100% = 10.48%}$