n_NaOH =0,3.1= 0,3 mol

CH₃COOC₆H₅ + 2NaOH → CH₃COONa + C₆H₅ONa + H₂O

 0,1                        0,2                 0,1                     0,1

m_rắn = m_CH3COONa + m_C6H5ONa + m_{NaOH dư}

= 0,1.82 + 0,1.116 + 0,1.40 = 23,8g

$n_{HCOOCH_2C_6H_5} = a(mol) ; n_{CH_3COOC_6H_5} = b(mol)$
Ta có : 

$136a + 136b = 27,2(1)$

$n_{KOH} = a + 2b = 0,1.2,5 = 0,25(2)$

Từ (1)(2) suy ra a = 0,15 ; b = 0,05

Y gồm : 

$n_{C_6H_5CH_2OH} = a = 0,15(mol)$
$n_{H_2O} = b = 0,05(mol)$

Bảo toàn khối lượng : $m = 27,2 + 0,25.56 - 0,15.108 -0,05.18 = 24,1(gam)$

$HCOOC_6H_5 + 2KOH \to HCOOK + C_6H_5OK + H_2O$

$n_{este}= \dfrac{12,2}{122} = 0,1(mol)$
$n_{KOH} = 0,2.2 = 0,4 > 2n_{este}$ nên KOH dư

$n_{H_2O} = n_{este} = 0,1(mol)$
Bảo toàn khối lượng : 

$m = 12,2 + 0,4.56 - 0,1.18 = 32,8(gam)$

Theo gt ta có: $n_{HCOOC_6H_5}=0,1(mol);n_{KOH}=0,4(mol)$

$\Rightarrow n_{H_2O}=0,1(mol)$

Bảo toàn khối lượng ta có: $m_{chatran}=32,8(g)$

Gọi $n_{HCOOCH_3} = n_{HCOOC_6H_5} = a(mol)$

Suy ra : 

$60a + 122a = 18,2 \Rightarrow a = 0,1$

$n_{NaOH} = 0,2.2 = 0,4(mol)$
$HCOOCH_3 + NaOH \to HCOONa + CH_3OH$
$HCOOC_6H_5 + 2NaOH \to HCOONa + C_6H_5ONa + H_2O$

$n_{H_2O} = n_{HCOOC_6H_5} = 0,1(mol)$
$n_{CH_3OH} = n_{HCOOCH_3} = 0,1(mol)$

Bảo toàn khối lượng : 

$m = 18,2 + 0,4.40 - 0,1.18 - 0,1.32 = 29,2(gam)$

\(n_{NaOH\left(dư\right)}=n_{HCl}=0.1\cdot1=0.1\left(mol\right)\)

\(\Rightarrow n_{NaOH\left(pư\right)}=0.6-0.1=0.5\left(mol\right)\)

\(n_{H_2O}=n_{NaOH}=0.5\left(mol\right)\)

\(BTKL:\)

\(m_{Muối}=24.2+0.5\cdot40-0.5\cdot18=35.2\left(g\right)\)

18) m=18,2g

1/ \(n_{KOH}=0,6.0,3=0,18\left(mol\right);n_{AlCl_3}=1,2.0,2=0,24\left(mol\right)\)

\(\Rightarrow n_{OH^-}=0,18\left(mol\right);n_{Al^{3+}}=0,24\left(mol\right)\)

\(Al^{3+}+3OH^-\rightarrow Al\left(OH\right)_3\)

\(0,06\leftarrow0,18\rightarrow0,06\)

\(\Rightarrow m_{Al\left(OH\right)_3}=0,06.78=4,68\left(g\right)\)

2/ \(n_{Al_2\left(SO_4\right)_3}=0,2\left(mol\right);n_{NaOH}=1,75.0,4=0,7\left(mol\right)\)

\(Al_2\left(SO_4\right)_3\rightarrow2Al^{3+}+3SO_4^{2-}\)

\(0,2---\rightarrow0,4---\rightarrow0,6\)

\(Al^{3+}+3OH^-\rightarrow Al\left(OH\right)_3\)

\(0,4--\rightarrow1,2--\rightarrow0,4\)

\(\Rightarrow m_{Al\left(OH\right)_3}=0,4.78=31,2\left(g\right)\)

Ủa ủa :D? Mấy câu này đăng lắm thế, 3 lần lận?