Bài 7 :

`m_{NaOH(A)}=400.10\%=40(g)`

`a,`

`n_{Na}=(9,2)/(23)=0,4(mol)`

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

            `0,4`                `->0,4`   `->0,2(mol)`

`=>m_{NaOH}=0,4.40=16(g)`

`m_{H_2}=0,2.2=0,4(g)`

`m_{dd (sau)}=m_{ddNaOH(A)}+m_{Na}-m_{H_2}`

                    `=400+9,2-0,4=408,8(g)`

`C\%_{NaOH(B)}=(16+40)/(408,8).100\%≈13,7\%`

`b,`Gọi `n_{Na_2O}=a`

PTHH:

`Na_2O+H_2O->2NaOH`

`a`                    `->2a`

`=>m_{NaOH}=40.2a=80a(g)`

Ta có:

`C\%_{NaOH(sau)}=(40+80a)/(62a+400)=12\%`

`=>a=0,11(mol)`

`=>m_{Na_2O}=0,11.62=6,82(g)`

Bài 8 :

Gọi `m_{SO_3}=a;m_{H_2SO_4(50\%)=b`

`=>m_{H_2SO_4}=50\%b=0,5b`

`n_{SO_3}=a/(80)(mol)`

`m_{H_2SO_4(sau)}=100.79\%=79(g)`

PTHH:`SO_3+H_2O->H_2SO_4(1)`

            `a/(80)`        `->a/(80)`

`=>m_{H_2SO_4(1)}=(98a)/(80)(g)`

Ta có :

`m_{H_2SO_4(sau)}=0,5b+(98)/(80)a=79`(*)

Ta lại có :`a+b=100`(**)

từ (*) và (**) ta suy ra :

`a=60;b=40`

Khối lượng `SO_3` cần dùng là `60(g)`

Bài 19 :

`n_{K_2O}=(47)/(94)=0,5(mol)`

PTHH: `K_2O+H_2O->2KOH(1)`

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

`=>m_{KOH(1)}=1.56=56(g)`

Gọi `m_{ddKOH(7,93\%)}=a`

`=>m_{KOH(dd)}=7,93\%a=0,0793a`

Ta có :

`C\%_{dd KOH(sau)}=(0,0793a+56)/(47+a)=21\%`

`=>a≈353(g)`

`#Ly`

$\text{Câu 7:}$

$a)$

$n_{Na}$ $=$ $\dfrac{9,2}{23}$ $=$ $0,4$ $(mol)$

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

Theo PT, $n_{NaOH}$ $=$ $n_{Na}$ $=$ $0,4$ $(mol)$

$m_{NaOH}$ $=$ $0,4.40=16$ $(g)$

Theo PT, $n_{H_2}$ $=$ $\dfrac{1}{2}n_{Na}$ $=$ $0,2$ $(mol)$

$m_{ddB}$ $=$ $400+9,2-0,2.2=408,8$ $(g)$

$m_{NaOH(A)}$ $=$ $\dfrac{400.10}{100}$ $=$ $40$ $(g)$

$m_{NaOH(B)}$ $=$ $40+16=56$ $(g)$

`C%_(NaOH(B)` $=$ $\dfrac{56}{408,8}$`.100%` $≈$ `13,697%`

$b)$

Gọi số $mol$ $Na_2O$ cần thêm là $a$ $(a>0)$

$m_{Na_2O}$ $=$ $62a$ $(g)$

PTHH: $Na_2O$ $+$ $H_2O$ `->` $2NaOH$

Theo PT, $n_{NaOH}$ $=$ $2n_{Na_2O}$ $=$ $2a$ $(mol)$

$m_{NaOH}$ $=$ $2a.40=80a$ $(g)$

Ta có:

$\dfrac{40+80a}{400+2a}$`.100%` $=$ `12%`

$⇒$ $a=0,1$

$m_{Na_2O}$ $=$ $0,1.62=6,2$ $(g)$

$\text{Câu 8:}$

Gọi số $mol$ $SO_3$ cần hòa tan là $a$ $(a>0)$

PTHH: $SO_3$ $+$ $H_2O$ `->` $H_2SO_4$

Theo PT, $n_{H_2SO_4}$ $=$ $n_{SO_3}$ $=$ $a$ $(mol)$

$m_{H_2SO_4}$ $=$ $98a$ $(mol)$

$m_{SO_3}$ $=$ $80a$ $(mol)$

`m_(H_2SO_4(79%)` $=$ $\dfrac{100.79}{100}$ $=$ $79$ $(g)$

Ta có:

$\dfrac{79-98a}{100-80a}$`.100%` $=$ `50%`

$⇒$ $a=0,725$

$m_{SO_3}$ $=$ $80.0,725=58$ $(g)$

$\text{Câu 9:}$

$n_{K_2O}$ $=$ $\dfrac{47}{94}$ $=$ $0,5$ $(mol)$

PTHH: $K_2O$ $+$ $H_2O$ `->` $2KOH$

Theo PT, $n_{KOH}$ $=$ $2n_{K_2O}$ $=$ $1$ $(mol)$

Gọi $x$ là khối lượng dd $KOH$ `7,93%` $(x>0)$

$m_{KOH}$ $=$ $\dfrac{x.79,3}{100}$ $=$ $0,0793x$ $(g)$

Ta có:

$\dfrac{0,0793x+0,1.56}{a+47}$`.100%` $=$ `21%`

$⇒$ $a=353$