Đáp án + Giải thích các bước giải:

 $m_{HCl(pứ)} = 73,73 : (100 + 1)\% = 73g$

$\Rightarrow n_{HCl(pứ)} = \dfrac{m}{M} = \dfrac{73}{35,5 + 1} = \dfrac{73}{36,5} = 2\text{ mol}$

PTHH:
$Zn + 2HCl \longrightarrow ZnCl_2 + H_2 \uparrow$
Vì $HCl$ dư nên tính theo $Zn$.

Theo PTHH:
$n_{Zn} = \dfrac{1}{2}n_{HCl(pứ)} = \dfrac{1}{2} . 2 = 1\text{ mol}$

$\Rightarrow m_{Zn} = n . M= 1. 65 = 65g$

Theo PTHH:
$n_{H_2} = n_{Zn} = 1\text{ mol}$
$\Rightarrow V_{H_2} =n . 22,4 = 1 . 22,4 = 22,4l$

Câu 6:
$n_{CaCO_3} = \dfrac{m}{M} = \dfrac{10}{40 + 12 + 3 . 16} = \dfrac{10}{100} = 0,1\text{ mol}$
$

$n_{CaO} = \dfrac{m}{M} = \dfrac{4,76}{40 + 16} = \dfrac{4,76}{56} = 0,085\text{ mol}$

PTHH:
$CaCO_3 \xrightarrow{t^o} CaO + CO_2 \uparrow$

Theo PTHH:
$n_{CaO(lt)} = n_{CaCO_3} = 0,1\text{ mol}$
$\Rightarrow H\% = \dfrac{n_{CaO(tt)}}{n_{CaO(lt)}} . 100\% = \dfrac{0,085}{0,1} . 100\% = 85\%$