Đáp án+Giải thích các bước giải:

 n$_{H2}$=$\frac{13,44}{22,4}$ = 0,6 (mol)

PTHH:

2Al+6HCl→2AlCl$_{3}$+3H$_{2}$  (1)

0,4← 1,2    ←        0,6   (mol)

Al$_{2}$O$_{3}$+6HCl→2AlCl$_{3}$+3H$_{2}$O  (2)

a,  Theo PTHH (1) ta suy ra : n$_{Al}$=0,4(mol)

⇒m$_{Al}$=0,4.27=10,8(g)

⇒m$_{Al2O3}$=31,2-10,8=20,4(g)

b,n$_{Al2O3}$=$\frac{20,4}{102}$ =0,2(mol)

⇒Theo PTHH (2)  : n$_{HCl (2)}$=6.n$_{Al2O3}$= 6.0,2=1,2(mol)

Theo PTHH (1) ta suy ra : n$_{HCl (1)}$=1,2(mol)

Do đó : n$_{HCl phản ứng}$ = 1,2+1,2 = 2,4(mol)

⇒V$_{HCl phản ứng}$ =$\frac{1}{2,4}$ =416(ml)

a/ $2Al+6HCl\to 2AlCl_3+3H_2$ (1)

$Al_2O_3+6HCl\to 2AlCl_3+3H_2O$ (2)

$n_{H_2}=\dfrac{13,44}{22,4}=0,6(mol)$

Theo PTHH(1): $n_{Al}=\dfrac{2}{3}n_{H_2}$

$→n_{Al}=0,4(mol)\\→m_{Al}=0,4.27=10,8g\\→m_{Al_2O_3}=20,4g$

Vậy $m_{Al}=10,8g;m_{Al_2O_3}=20,4$

b/ Theo PTHH(1): $n_{HCl}=3n_{Al}$

$→n_{HCl}=1,2(mol)$

$n_{Al_2O_3}=\dfrac{20,4}{102}=0,2(mol)$

Theo PTHH(2): $n_{HCl}=6n_{Al_2O_3}$

$→n_{HCl}=1,2(mol)$

$∑n_{HCl}=1,2+1,2=2,4(mol)\\→V_{HCl}=\dfrac{n}{C_M}=\dfrac{2,4}{1}=2,4l$

Vậy $V_{HCl}=2,4l$