Đáp án + Giải thích các bước giải:

 Bài 4:
a) $200ml = 0,2l$

$\Rightarrow n_{KNO_3} = C_M . V = 1,5 . 0,2 = 0,3\text{ mol}$
$\Rightarrow m_{KNO_3} = n . M = 0,3 . (39 + 14 + 3 . 16) = 0,3 . 101 = 30,3g$

b) $m_{NaOH} = \dfrac{C\% . m_{dd}}{100\%} = \dfrac{10\% . 250}{100\%} = 25g$

$\Rightarrow n_{NaOH} = \dfrac{m}{M} = \dfrac{25}{23 + 16 + 1} = \dfrac{25}{40} = 0,625\text{ mol}$

Bài 5:

a) PTHH:
$Zn + 2HCl \longrightarrow ZnCl_2 + H_2 \uparrow$
b) $n_{Zn} = \dfrac{m}{M} = \dfrac{32,5}{65} = 0,5\text{ mol}$

Theo PTHH:
$n_{ZnCl_2} = n_{H_2} = n_{Zn} = 0,5\text{ mol}$
$\Rightarrow m_{ZnCl_2}  = n . M = 0,5 . (65 + 2 . 35,5) = 0,5 . 136 = 68g$

$V_{H_2} =n . 22,4 = 0,5 . 22,4 = 11,2l$

$4)a$

Đổi $200(ml)=0,2(l)$

`=>n_(KNO_3)=0,2.1,5=0,3(mol)`

`=>m_(KNO_3)=0,3.101=30,3(g)`

$b$

`m_(ctNaOH)=(250.10)/100=25(g)`

`=>n_(NaOH)=25/40=0,625(mol)`

$5)$

`n_(Zn)=(32,5)/65=0,5(mol)`

$a.PTHH:$

$Zn+2HCl \to ZnCl_2+H_2$

$b.$

Theo `PT,n_(ZnCl_2)=n_(Zn)=0,5(mol)`

`=>m_(ZnCl_2)=0,5.136=68(g)`

Theo `PT,n_(H_2)=n_(Zn)=0,5(mol)`

`=>V_(H_2)(đktc)=0,5.22,4=11,2(l)`