Đáp án + Giải thích các bước giải:

Câu 11:

$a\big)$

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

$Zn+2HCl\to ZnCl_2+H_2$

Vì $n_{Zn}<\frac{n_{HCl}}{2}\to HCl$ dư

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

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

$b\big)$

$m_{dd\,HCl}=0,4.36,5+50=64,6(g)$

$m_{dd\,spu}=6,5+64,6-0,1.2=70,9(g)$

Theo PT: $n_{HCl(dư)}=0,4-0,1.2=0,2(mol);n_{ZnCl_2}=n_{Zn}=0,1(mol)$

$\to\begin{cases} C\%_{HCl(dư)}=\frac{0,2.36,5}{70,9}.100\%\approx 10,3\%\\ C\%_{ZnCl_2}=\frac{0,1.136}{70,9}.100\%\approx 19,18\% \end{cases}$

Câu 12:

$n_{Zn}=\frac{2,6}{65}=0,04(mol)$

$n_{HCl}=\frac{32,85}{36,5}=0,9(mol)$

$Zn+2HCl\to ZnCl_2+H_2$

Vì $n_{Zn}<\frac{n_{HCl}}{2}\to HCl$ dư

Theo PT: $n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,04(mol)$

$\to\begin{cases} m_{ZnCl_2}=0,04.136=5,44(g)\\ V_{H_2}=0,04.22,4=0,896(l) \end{cases}$

@$\text{Ly}$

$\text{Câu 11:}$

$a)$

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

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

Ta có: $\dfrac{0,1}{1}$ $<$ $\dfrac{0,4}{2}$

$⇒$ $Zn$ pư hết, $HCl$ dư nên tính theo $n_{Zn}$

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

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

$b)$

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

$n_{HCl(dư)}$ $=$ $0,4-0,2=0,2$ $mol$

$m_{HCl(dư)}$ $=$ $0,2.36,5=7,3$ $(g)$

$m_{HCl(bđ)}$ $=$ $0,4.36,5=14,6$ $(g)$

$m_{ddHCl(bđ)}$ $=$ $m_{HCl(bđ)}$ $+$ $m_{H_2O}$ $=$ $14,6+50=64,6$ $(g)$

$m_{dd(spư)}$ $=$ $6,5+64,6-(0,1.2)=70,9$ $(g)$

`C%_(HCl(spư)` $=$ $\dfrac{7,3}{70,9}$`.100%` $≈$ $10,296$`%`

Theo PT, $n_{ZnCl_2}$ $=$ $n_{Zn}$ $=$ $0,1$ $mol$

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

`C%_(ZnCl_2` $=$ $\dfrac{13,6}{70,9}$`.100%` $≈$ $19,18$`%`

$\text{Câu 12:}$

$n_{Zn}$ $=$ $\dfrac{2,6}{65}$ $=$ $0,04$ $mol$

$n_{HCl}$ $=$ $\dfrac{32,85}{36,5}$ $=$ $0,9$ $mol$

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

Ta có: $\dfrac{0,04}{1}$ $<$ $\dfrac{0,9}{2}$

$⇒$ $Zn$ pư hết, $HCl$ dư nên tính theo $n_{Zn}$

Theo PT, $n_{H_2}$ $=$ $n_{ZnCl_2}$ $=$ $n_{Zn}$ $=$ $0,04$ $mol$

$V_{H_2}$ $=$ $0,04.22,4=0,896$ $(l)$

$m_{ZnCl_2}$ $=$ $0,04.136=5,44$ $(g)$