Đáp án + Giải thích các bước giải:

$a)$

$Zn+2HCl\to ZnCl_2+H_2$

$b)$

$n_{Zn}=\dfrac{6,5}{65}=0,1(mol)$

$n_{H_2}=n_{Zn}=0,1(mol)$

$V_{H_2}=0,1×22,4=2,24(l)$

$c)$

$n_{HCl}=2n_{Zn}=0,2(mol)$

$C\%_{HCl}=\dfrac{0,2×36,5}{200}×100\%=3,65\%$

$d)$

$m_{dd\ sau}=6,5+200-0,1×2=206,3(g)$

$n_{ZnCl_2}=n_{Zn}=0,1(mol)$

$C\%_{ZnCl_2}=\dfrac{0,1×136}{206,3}×100\%=6,59\%$

@$\text{Ly}$

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

$b)$ $n_{Zn}$ $=$ $\dfrac{6,5}{65}$ $=$ $0,1$ $mol$

Theo PT, $n_{H_2}$ $=$ $n_{Zn}$ $=$ $0,1$ $mol$

$V_{H_2}$ $=$ $0,1.22,4=2,24$ $(l)$

$c)$ Tương tự, $n_{HCl}$ $=$ $2.n_{Zn}$ $=$ $0,2$ $mol$

$m_{HCl}$ $=$ $36,5.0,2=7,3$ $(g)$

`C%_(HCl` $=$ $\dfrac{7,3}{200}$`.100%` $=$ $3,65$`%`

$d)$ Ta có: $n_{ZnCl_2}$ $=$ $n_{Zn}$ $=$ $0,1$ $mol$

$m_{ZnCl_2}$ $=$ $0,1.136=13,6$ $(g)$

$m_{ddZnCl_2}$ $=$ $6,5+200-(0,1.2)=206,3$ $(g)$

`C%_(ZnCl_2` $=$ $\dfrac{13,6}{206,3}$`.100%` $≈$ $6,59$`%`