Đáp án:

Giải thích các bước giải:

$a/$

$2Na + 2H_2O \to 2NaOH + H_2$

$CaO + H_2O \to Ca(OH)_2$
$b/$

$n_{H_2} = \dfrac{1,12}{22,4}  = 0,05(mol)$

Theo PTHH :

$n_{Na} = n_{NaOH} = 2n_{H_2} = 0,05.2 = 0,1(mol)$

$\to n_{CaO} = \dfrac{7,9 - 0,1.23}{56} = 0,1(mol)$

$\to n_{Ca(OH)_2} = n_{CaO} = 0,1(mol)$

Sau phản ứng :

$\to V_{dd} = \dfrac{m}{D} = \dfrac{500}{1,2} = 416(ml) = 0,416(lít)$
Vậy :

$C_{M_{NaOH}} = \dfrac{0,1}{0,416} = 0,24M$

$C_{M_{Ca(OH)_2}} = \dfrac{0,1}{0,423} = 0,24M$

`n_(H_2)=\frac{1,12}{22,4}=0,05(mol)`

`a,PTHH`

`2Na+2H_2O-> 2NaOH+H_2`

   `0,1`                                       `0,05`

`CaO+H_2O-> Ca(OH)_2`

`b, V_(dd)=\frac{m}{D}=\frac{500}{1,2}=\frac{1250}{3}(ml)=\frac{5}{12} (l)`

`m_(Na)=0,1.23=2,3(g)`

`=> m_(CaO)=7,9-2,3=5,6(g)`

`=> n_(CaO)=\frac{5,6}{56}=0,1(mol)`

`=> C_(MNa)=\frac{0,1}{\frac{5}{12}}=0,24M`

`=> C_(MCaO)=\frac{0,1}{\frac{5}{12}}=0,24M`