Đáp án+ Giải thích các bước giải:

`PT : Zn + 2HCl -> ZnCl_2 + H_2(1)`

`n_{Zn} = m/M = 13/65 = 0,2(mol)`

`a)`

Theo pt `(1)` : `n_{H_2} = n_{Zn} = 0,2(mol)`

`->` `V_{H_2(đktc)} = n.22,4 = 0,2.22,4 = 4,48(l)`

`b)`

Theo pt `(1)` : `n_{HCl} = 2n_{Zn} = 2.0,2 = 0,4(mol)`

`->` `m_{HCl} = n.M = 0,4.36,5 = 14,6(g)`

`c)`

$m_{\text{dd HCl}}$ = `\frac{m_{ct}.100%}{C%}` = `\frac{14,6.100%}{15%}` `= 97,3(g)`

`d)`

Theo pt `(1)` : `n_{ZnCl_2} = n_{Zn} = 0,2(mol)`

$m_{\text{dd spứ}}$ = `m_{Zn} + m_{HCl} - m_{H_2} = 13 + 97,3 - (0,2.2) = 109,9(g)`

`-> C%`$_{\text{dd $ZnCl_2$}}$ = $\dfrac{m_{ct}}{m_{dd}}$.`100(%)` = $\dfrac{0,2.136}{109,9}$.`100(%)` $\approx$ `24,75(%)`

$\text{#Cyrus}$

Đáp án + Giải thích các bước giải:

$a\big)$

$n_{Zn}=\frac{13}{65}=0,2(mol)$

$Zn+2HCl\to ZnCl_2+H_2$

Theo PT: $n_{H_2}=n_{Zn}=0,2(mol)$

$\to V_{H_2(đktc)}=0,2.22,4=4,48(l)$

$b\big)$

Theo PT: $n_{HCl}=2n_{Zn}=0,4(mol)$

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

$c\big)m_{dd\,HCl}=\frac{14,6}{15\%}\approx 97,33(g)$

$d\big)$

Theo PT: $n_{ZnCl_2}=n_{Zn}=0,2(mol)$

$\to C\%_{ZnCl_2}=\frac{0,2.136}{13+97,33-0,2.2}.100\%\approx 24,74\%$