`1)`

`n_{Na}=(23)/(23)=1(mol)`

PTHH:

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

`1`                        `->1`      `->0,5(mol)`

Khối lượng dung dịch sau phản ứng :

`m_{dd(sau)}=23+100-0,5.2=122(g)`         

`m_{NaOH}=1.40=40(g)`           

`C\%_{NaOH}=(40)/(122).100\%≈32,79\%`

`2)`

`n_{CaCO_3}=(15)/(100)=0,15(mol)`

`n_{HCl}=0,1.0,2=0,02(mol)`

PTHH:

`CaCO_3+2HCl->CaCl_2+CO_2+H_2O`

Ta có tỉ lệ :

`(n_{CaCO_3})/1 = (0,15)/1 > (n_{HCl})/2 = (0,02)/2`

`=>` Sau phản ứng `CaCO_3` dư, `HCl` hết

`a,`

`n_{CO_2}= 1/2 n_{HCl } = 1/2 . 0,02=0,01(mol)`

`=>V_{CO_2}=0,01.22,4=0,224(lít)`

`b,` Đổi `100ml=0,1` lít

`n_{CaCl_2}= 1/2 n_{HCl} = 1/2 . 0,02=0,01(mol)`

`=>C_{M CaCl_2}=(0,01)/(0,1)=0,1(M)`

@$\text{Ly}$

Câu 1:

$n_{Na}$ $=$ $\dfrac{m}{M}$ $=$ $\dfrac{23}{23}$ $=$ $1$ $mol$

PTHH: $2Na$ $+$ $2H_2O$ `->` $2NaOH$ $+$ $H_2↑$

Theo PT, $n_{NaOH}$ $=$ $n_{Na}$ $=$ $1$ $mol$

$m_{NaOH}$ $=$ $1.40=40$ $(g)$

Tương tự, $n_{H_2}$ $=$ $0,5n_{Na}$ $=$ $0,5$ $mol$

$→$ $m_{H_2}$ $=$ $n.M$ $=$ $0,5.2=1$ $(g)$

$m_{ddNaOH}$ $=$ $m_{Na}$ $+$ $m_{H_2O}$ $-$ $m_{H_2}$ $=$ $23+100-1=122$ $(g)$

`C%_(NaOH` $=$ $\dfrac{m_{NaOH}}{m_{ddNaOH}}$`.100%` $=$ $\dfrac{40}{122}$`.100%` $≈$ $32,79$`%`

Câu 2:

$a)$

$n_{CaCO_3}$ $=$ $\dfrac{m}{M}$ $=$ $\dfrac{15}{100}$ $=$ $0,15$ $mol$

Đổi $100$ $ml$ $=$ $0,1$ $l$

$n_{HCl}$ $=$ $C_M.V$ $=$ $0,2.0,1=0,02$ $mol$

PTHH: $CaCO_3$ $+$ $2HCl$ `->` $CaCl_2$ $+$ $CO_2↑$ $+$ $H_2O$

Xét tỉ lệ: $\dfrac{n_{CaCO_3}}{1}$ $=$ $\dfrac{0,15}{1}$ Và $\dfrac{n_{HCl}}{2}$ $=$ $\dfrac{0,02}{2}$

Mà $\dfrac{0,15}{1}$ $>$ $\dfrac{0,02}{2}$

$⇒$ $CaCO_3$ còn dư, $HCl$ phản ứng hết nên tính theo $n_{HCl}$ 

Theo PT, $n_{CO_2}$ $=$ $0,5n_{HCl}$ $=$ $0,01$ $mol$

$V_{CO_2}$ $=$ $n.22,4=0,01.22,4=0,224$ $(l)$

$b)$ Tương tự, $n_{CaCl_2}$ $=$ $0,5n_{HCl}$ $=$ $0,01$ $mol$

`C_MNaCl_2` $=$ $\dfrac{n}{V}$ $=$ $\dfrac{0,01}{0,1}$ $=$ $0,1$ $M$