n_Al(OH)3=\(\dfrac{23.4}{78}\)=0.3(mol)

n_NaOH=0.2*2=0.4(mol)

Al(OH)₃+NaOH---->NaAlO₂+2H₂O

mol:\(\dfrac{0.3}{1}\)<\(\dfrac{0.4}{1}\) => NaOH dư

Al(OH)₃+NaOH---->NaAlO₂+2H₂O

mol:0.3--->0.3--------->0.3 (mol)

m_NaAlO2=82*0.3=24.6(g)

n_{NaOH dư}=0.4-0.3=0.1(mol)

m_{NaOH thu được}=40*0.1=4(g)

Đáp án A

Từ %O=> M_X= 89 => n_X=0.3 mol , n_NaOH = 0.4 mol

BTKL : 26.7+ 0.4*40=m +0.3*18 => m =37.3 gam

Đáp án là A

Ta có n_H2SO4 = 0,2 . 1,5 = 0,3 ( mol )

n_Ba(OH)2 = 0,3 . 0,8 = 0,24 ( mol )

H₂SO₄ + Ba(OH)₂ → BaSO₄ + 2H₂O

0,3...........0,24

⇒Lập tỉ số 

1) \(n_{Al\left(OH\right)_3}=\dfrac{0,78}{78}=0,01\left(mol\right)\)

PTHH: \(Al_2\left(SO_4\right)_3+6NaOH\rightarrow3Na_2SO_4+2Al\left(OH\right)_3\) 

                                   0,03<----------------------0,01

=> n_{NaOH min} = 0,03 (mol)

=> \(C_{M\left(NaOH\right)}=\dfrac{0,03}{0,2}=0,15M\)

2) \(n_{Al_2O_3}=\dfrac{5,1}{102}=0,05\left(mol\right)\)

\(n_{Al_2\left(SO_4\right)_3}=0,3.0,25=0,075\left(mol\right)\)

PTHH: \(6NaOH+Al_2\left(SO_4\right)_3\rightarrow3Na_2SO_4+2Al\left(OH\right)_3\)

           0,45<------0,075-------------------------->0,15

            \(NaOH+Al\left(OH\right)_3\rightarrow NaAlO_2+2H_2O\)

             0,05<----0,05

            \(2Al\left(OH\right)_3\underrightarrow{t^o}Al_2O_3+3H_2O\)

              0,1<-------0,05

=> n_{NaOH max} = 0,5 (mol)

=> \(V_{dd}=\dfrac{0,5}{2}=0,25\left(l\right)=250\left(ml\right)\)

3)

\(n_{KOH\left(1\right)}=0,15.1,2=0,18\left(mol\right)\)

\(n_{Al\left(OH\right)_3\left(1\right)}=\dfrac{4,68}{78}=0,06\left(mol\right)\)

\(n_{AlCl_3}=0,1.x\left(mol\right)\)

Do khi cho KOH tác dụng với dd Y xuất hiện kết tủa

=> Trong Y chứa AlCl₃ dư

PTHH: \(3KOH+AlCl_3\rightarrow3KCl+Al\left(OH\right)_3\)

           0,18---->0,06----------------->0,06

\(n_{KOH\left(2\right)}=0,175.1,2=0,21\left(mol\right)\)

\(n_{Al\left(OH\right)_3\left(2\right)}=\dfrac{2,34}{78}=0,03\left(mol\right)\)

PTHH: \(3KOH+AlCl_3\rightarrow3KCl+Al\left(OH\right)_3\)

   (0,3x-0,18)<--(0,1x-0,06)------->(0,1x-0,06)

            \(KOH+Al\left(OH\right)_3\rightarrow KAlO_2+2H_2O\)

     (0,1x-0,09)<-(0,1x-0,09)

=> \(\left(0,3x-0,18\right)+\left(0,1x-0,09\right)=0,21\)

=> x = 1,2

a) nCH3COOH= 0,4(mol)

PTHH: CH3COOH + NaOH -> CH3COONa + H2O

0,4____________0,4(mol)

=> mNaOH=0,4. 40=16(g)

b) nCH3COOH= 1(mol)

nC2H5OH= 100/46= 50/23(mol)

Vì : 1/1< 50/23 :1

=> C2H5OH dư, CH3COOH hết, tính theo nCH3COOH.

PTHH: CH3COOH + C2H5OH \(⇌\) CH3COOC2H5 + H2O (đk: H⁺ , nhiệt độ)

Ta có: nCH3COOC2H5(thực tế)= 0,625(mol)

Mà theo LT: nCH3COOC2H5(LT)= nCH3COOH=1(mol)

=>H= (0,625/1).100=62,5%

Ý bạn là tìm \(V_{dd_{HCl}}\) đúng không?

\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\\ AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3\downarrow+3NaCl\\ \Rightarrow n_{AlCl_3}=\dfrac{1}{3}n_{NaOH}=0,1\left(mol\right)\\ \Rightarrow n_{HCl}=3n_{AlCl_3}=0,3\left(mol\right)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,3}{2}=0,15\left(l\right)\)