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

Em ko có tiền nên mong a , cj tham khảo(đây là 500 g dung dịch nên mong a, cj sửa 1 chút ạ)

a)

$BaCl_2 + H_2SO_4 \to BaSO_4 + 2HCl$
$n_{BaCl_2} = 0,1 < n_{H_2SO_4} = 0,2$ nên $H_2SO_4$ dư

$n_{BaSO_4} = n_{BaCl_2} = 0,1(mol)$
$m_{BaSO_4} = 0,1.233 = 23,3(gam)$

b)

A gồm :

$HCl : 0,1.2 = 0,2(mol)$
$H_2SO_4\ dư : 0,2 - 0,1 = 0,1(mol)$
$V_{dd} = 0,1 + 0,1= 0,2(lít)$

$C_{M_{HCl}} = \dfrac{0,2}{0,2} = 1M$
$C_{M_{H_2SO_4}} = \dfrac{0,1}{0,2} = 0,5M$

c)

$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{NaOH} = 2n_{H_2SO_4\ dư} = 0,2(mol)$
$m_{dd\ NaOH} = \dfrac{0,2.40}{15\%} = 53,33(gam)$

Xem lại đề !

\(m_{ct}=\dfrac{19,6.200}{100}=39,2\left(g\right)\)

\(n_{H2SO4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)

\(m_{ct}=\dfrac{5,2.200}{100}=10,4\left(g\right)\)

\(n_{BaCl2}=\dfrac{10,4}{208}=0,05\left(mol\right)\)

Pt : \(H_2SO_4+BaCl_2\rightarrow2HCl+BaSO_4|\)

            1             1             2             1

          0,4           0,05        0,1         0,05

a) Lập tỉ số so sánh: \(\dfrac{0,4}{1}>\dfrac{0,05}{1}\)

             ⇒ H₂SO₄ dư , BaCl₂ phản ứng hết

             ⇒ Tính toán dựa vào số mol của BaCl₂

\(n_{BaSO4}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)

⇒ \(m_{BaSO4}=0,05.233=11,65\left(g\right)\)

b) \(n_{HCl}=\dfrac{0,05.2}{1}=0,1\left(mol\right)\)

⇒ \(m_{HCl}=0,1.36,5=3,65\left(g\right)\)

\(n_{H2SO4\left(dư\right)}=0,4-0,05=0,35\left(mol\right)\)

⇒ \(m_{H2SO4\left(dư\right)}=0,35.98=34,3\left(g\right)\)

\(m_{ddspu}=200+200-11,65=388,35\left(g\right)\)

\(C_{ddHCl}=\dfrac{3,65.100}{388,35}=0,94\)⁰/₀

\(C_{ddH2SO4\left(dư\right)}=\dfrac{34,3.100}{388,35}=8,83\)⁰/₀

 Chúc bạn học tốt

\(a.n_{H_2SO_4}=\dfrac{200.19,6\%}{98}=0,4\left(mol\right)\\ n_{BaCl_2}=\dfrac{200.5,2\%}{208}=0,05\left(mol\right)\\ BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ Vì:\dfrac{0,05}{1}< \dfrac{0,4}{1}\\ \Rightarrow H_2SO_4dư\\ n_{BaSO_4}=n_{BaCl_2}=0,05\left(mol\right)\\ \Rightarrow m_{\downarrow}=m_{BaSO_4}=0,05.233=11,65\left(g\right)\\ b.m_{ddsau}=200+200-11,65=388,35\left(g\right)\\ C\%_{ddHCl}=\dfrac{0,05.2.36,5}{388,35}.100\approx0,94\%\\ C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{\left(0,4-0,05\right).98}{388,35}.100\approx8,832\%\)

Đáp án D