Giải thích các bước giải:

a. Gọi hóa trị của R là n

$2R+2nHCl\to 2RCl_n+nH_2$

$\Delta_m=m_R-m_{H_2}$

$⇒m_{H_2}=21,6-19,2=2,4\ (gam)$

$⇒n_{H_2}=\dfrac{2,4}2=1,2\ (mol)$

BTNT H: $n_{HCl}=2n_{H_2}=2,4\ (mol)$

$⇒m_{RCl_n}=21,6+2,4.35,5=106,8\ (gam)$

b.

Theo PTHH: $n_R=\dfrac 2n.n_{H_2}=\dfrac{2,4}n\ (gam/mol)$

$⇒M_R=\dfrac{21,6}{\dfrac{2,4}n}=9n\ (gam/mol)$

mà n là hóa trị của kim loại nên $1\leqslant n\leqslant 3$

$⇒\begin{cases}n=3\\ M_R=27\ (Al)\end{cases}$

PTHH:

$2Al+6HCl\to 2AlCl_3+3H_2$

BTKL:

$m_R+m_{dd\ HCl}=m_{dd\ T}+m_{H_2}$

$⇒m_{dd\ T}=21,6+100-2,4=119,2\ (gam)$

$⇒C\%_{dd\ T}=\dfrac{106,8}{119,2}\cdot 100\%=89,59\%$

$n_{HCl}=2,4\ (mol)$

$⇒C\%_{dd\ HCl}=\dfrac{2,5.36,5}{100}\cdot 100\%=91,25\%$

c.

BTNT Al: $n_{AlCl_3}=n_{Al}=\dfrac{21,6}{27}=0,8\ (mol)$

$n_{NaOH}=0,13.2=0,26\ (mol)$

PTHH:

$AlCl_3+3NaOH\to 3NaCl+Al(OH)_3\\ \dfrac{13}{150}\leftarrow\ \ \ 0,26\ \ \ \ \to \ \ \ \ 0,26\to \dfrac{13}{150}$

Xét: $0,8>\dfrac{0,26}3⇒$ NaOH hết

$⇒n_{Al(OH)_3}=\dfrac{13}{150}\ (mol)$

$⇒m_{Al(OH)_3}=\dfrac{13}{150}.78=6,76\ (gam)$

$n_{AlCl_3\ du}=0,8-\dfrac{13}{150}=\dfrac{107}{150}\ (mol)$

$⇒m_{AlCl_3\ du}=133,5\cdot \dfrac{107}{150}=95,23\ (gam)$

$n_{NaCl}=0,26\ (mol)⇒m_{NaCl}=0,26.58,5=15,21\ (gam)$

d.

$2Al(OH)_3\xrightarrow{t^\circ}Al_2O_3+3H_2O$

BTNT Al: $n_{Al_2O_3}=\dfrac 12\cdot n_{Al(OH)_3}=\dfrac{13}{300}\ (mol)$

$⇒m_{Al_2O_3}=102\cdot \dfrac{13}{300}=4,42\ (gam)$