giải hộ bài hóa với các bro thank nhìu ạ

@$\text{Ly}$

$a)$ $n_{Al}$ $=$ $\dfrac{5,4}{27}$ $=$ $0,2$ $mol$

PTHH: $2Al$ $+$ $6HCl$ `->` $2AlCl_3$ $+$ $3H_2↑$

$b)$ Theo PT, $n_{H_2}$ $=$ $\dfrac{3}{2}.n_{Al}$ $=$ $0,3$ $mol$

$V_{H_2}$ $=$ $0,3.22,4=6,72$ $(l)$

$c)$ Tương tự, $n_{AlCl_3}$ $=$ $n_{Al}$ $=$ $0,2$ $mol$

$m_{AlCl_3}$ $=$ $0,2.133,5=26,7$ $(g)$

$d)$ Theo PT, $n_{HCl}$ $=$ $3.n_{Al}$ $=$ $0,6$ $mol$

$m_{HCl}$ $=$ $0,6.36,5=21,9$ $(g)$

$m_{ddHCl}$ $=$ $\dfrac{21,9.100}{10}$ $=$ $219$ $(g)$