Đáp án  + Giải thích các bước giải:

 a) $n_{Zn} = \dfrac{m}{M} = \dfrac{6,5}{65} = 0,1(mol)$
$n_{HCl} = \dfrac{m}{M} = \dfrac{36,5}{35,5 + 1} = \dfrac{36,5}{36,5} = 1(mol)$

PTHH:
$Zn + 2HCl \longrightarrow ZnCl_2 + H_2 \uparrow$
Ta có:
$n_{Zn} < \dfrac{n_{HCl}}{2}\bigg(0,1 < \dfrac{1}{2}\bigg)$
$\Rightarrow HCl$ dư, tính theo $Zn$

Theo PTHH:
$n_{HCl(pứ)} = 2n_{Zn} = 2 . 0,1 = 0,2(mol)$
$\Rightarrow n_{HCl(\text{dư})} = 1 - 0,2 = 0,8(mol)$
$\Rightarrow m_{HCl(\text{dư})} = n . M = 0,8 . (35,5 + 1) = 0,8 . 36,5 = 29,2(g)$

b) Theo PTHH:
$n_{H_2} = n_{Zn} = 0,1(mol)$

$\Rightarrow V_{H_2} = n . 22,4 = 0,1 . 22,4 = 2,24(l)$
c) PTHH:
$2H_2 + O_2 \xrightarrow{t^o} 2H_2O$

Theo PTHH:
$n_{O_2} =\dfrac{1}{2}n_{H_2}= \dfrac{1}{2} . 0,1 = 0,05(mol)$
$\Rightarrow m_{O_2} = n . M = 0,05 . 2 . 16 = 1,6(g)$

$\color{red}{\text{#Vexi}}$

`#Ly`

$a)$

`n_(Zn` `=` `(6,5)/(65)=0,1(mol)`

`n_(HCl` `=` `(36,5)/(36,5)=1(mol)`

PTHH: $Zn$ $+$ $2HCl$ `->` $ZnCl_2$ $+$ $H_2$

Do: `0,1<(1)/(2)` `->` $HCl$ dư

Theo PT: `n_(HCl(pư)` `=` `2n_(Zn` `=` `0,2(mol)`

`n_(HCl(dư)` `=` `1-0,2=0,8(mol)`

`m_(HCl(dư)` `=` `0,8.36,5=29,2(g)`

$b)$

Theo PT: `n_(H_2` `=` `n_(Zn` `=` `0,1(mol)`

`V_(H_2` `=` `0,1.22,4=2,24(l)`

$c)$

PTHH: $2H_2$ $+$ $O_2$ $\xrightarrow[]{t^o}$ $2H_2O$

Theo PT: `n_(O_2` `=` `(1)/(2)n_(H_2` `=` `0,05(mol)`

`m_(O_2` `=` `0,05.32=1,6(g)`