4/

25ml=0,025l.

$n_{NaOH}=1.0,025=0,025(mol)$

$3NaOH + FeCl_3 \xrightarrow{} Fe(OH)_3 + 3NaCl$

$0,025$ : $\frac{1}{120}$ : $\frac{1}{120}$ : $0,025$ (mol)

$m_{Fe(OH)_{3}}=\dfrac{1}{120}.(56+17.3)≈0,891 (g)$

(mình làm tắt, vì nNaOH/3 < nFeCl3/1 nên ta lấy mol của NaOH, bạn nếu mún thì lập tỉ lệ mol để giáo viên ko trừ điểm nha)

5/ 25ml=0,025l

$n_{HCl}=1.0,025=0,025(mol)$

$n_{NaHCO_3}=1.0,025=0,025(mol)$

$HCl + NaHCO_3 \xrightarrow{} NaCl + H_2O + CO_2$

$0,025$ : $0,025$ : $0,025$ : $0,025$ (mol)

$V_{CO_{2}}=22,4.0,025=5,6(l)$

6/

100ml=0,1l

$n_{H_{2}SO_{4}}=1.0,1=0,1 (mol)$

$n_{H^{+}}=0,1.2=0,2 (mol)$ ( vì $H_{2}SO_{4}$ điện li ra 2 $H^{+}$)

$n_{NaOH}=n_{OH^{-}}=1.V=V (mol)$

$H^{+} + OH^{-} \xrightarrow{} H_2O$

1 : 1 : 1  

=> $n_{OH^{-}} = n_{H^{+}}$ 

<=>$n_{OH^{-}} =V = 0,2$ (mol)  ( V này là ẩn )
$V_{NaOH}=\frac{0,2}{1}=0,2(l)=200ml$

Như vậy giá trị của V = 200

7/ 
100ml = 0,1l, 200ml=0,2l

$n_{NaOH}=n_{OH^{-}}=0,01.0,1=0,001 (mol)$

$n_{HCl}=n_{H^{+}}=0,03.0,1=0,003(mol)$

$V_{dd}sau=0,2 (l)$

$H^{+} + OH^{-} \xrightarrow{} H_2O$

0,001 : 0,001 : 0,001 (mol)
=> $n_{H^{+}}$dư$=0,003-0,001=0,002 (mol)$

$[H^{+}]=\frac{0,002}{0,2}=0,01(M)$

$pH=-lg(0,01)=2$

Chúc bạn học tốt!

4.

$n_{NaOH}=0,025(mol)$

$FeCl_3+3NaOH\to Fe(OH)_3+3NaCl$

$\Rightarrow n_{Fe(OH)_3}=\dfrac{0,025}{3}=\dfrac{1}{120}(mol)$

$\to m=107.\dfrac{1}{120}=0,89g$

5.

$n_{HCl}=0,025(mol)$

$n_{NaHCO_3}=0,025(mol)$

$HCl+NaHCO_3\to NaCl+CO_2+H_2O$

$\Rightarrow n_{CO_2}=0,025(mol)$

$\to V=0,025.22,4=0,56l$

6.

$n_{H_2SO_4}=0,1(mol)$

$2NaOH+H_2SO_4\to Na_2SO_4+2H_2O$

$\Rightarrow n_{NaOH}=0,1.2=0,2(mol)$

$\to V=\dfrac{0,2}{1}=0,2l=200ml$

7.

$n_{OH^-}=n_{NaOH}=0,1.0,01=0,001(mol)$

$n_{H^+}=n_{HCl}=0,1.0,03=0,003(mol)$

$\Rightarrow$ dư $0,003-0,001=0,002$ mol $H^+$

$[H^+]=\dfrac{0,002}{0,2}=0,01M$

$\to pH=-\lg0,01=2$