$\text{Câu a:}$
$\text{$n_{Al}$ = 16.2/27 = 0.6 mol}$
$\textit{2Al + 6HCl → 2$AlCl_{3}$ + 3$H_{2}$↑}$
$\text{$n_{H_{2}}$ = 3/2$n_{Al}$ = 0.9 mol ⇒ $V_{H_{2}}$ = 0.9*22.4 = 20.16 lít}$
$\text{Câu b:}$
$\text{$n_{HCl}$ = 3$n_{Al}$ = 1.8 mol ⇒ $V_{HCl}$ = 1.8/(900/1000) = 2 M}$
$\text{Câu c:}$
$\text{$n_{CuO}$ = 52/80 = 0.65 mol}$
$\textit{CuO + $H_{2}$ → Cu + $H_{2}O$}$
$\text{Ta có tỉ lệ: $\frac{0.65}{1}$ < $\frac{0.9}{1}$ ⇒ $n_{H_{2}}$ dư}$
$\text{$n_{Cu}$ = $n_{CuO}$ = 0.65 mol}$
$\text{⇒ $m_{cr}$ = $m_{Cu}$ = 0.65*64 = 41.6 g}$