a)

\(n_{CuCl_2}=0,1.1,5=0,15\left(mol\right)\)

\(n_{Ca\left(OH\right)_2}=0,3.1=0,3\left(mol\right)\)

PTHH: CuCl₂ + Ca(OH)₂ --> Cu(OH)₂ + CaCl₂

Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,3}{1}\) => CuCl₂ hết, Ca(OH)₂ dư

PTHH: CuCl₂ + Ca(OH)₂ --> Cu(OH)₂\(\downarrow\) + CaCl₂

_____0,15---->0,15-------->0,15---------->0,15

=> \(\left\{{}\begin{matrix}C_{M\left(Ca\left(OH\right)_2dư\right)}=\dfrac{0,3-0,15}{0,1+0,3}=0,375M\\C_{M\left(CaCl_2\right)}=\dfrac{0,15}{0,1+0,3}=0,375M\end{matrix}\right.\)

b) Khối lượng giảm = khối lượng H₂O sinh ra

\(n_{H_2O}=\dfrac{0,9}{18}=0,05\left(mol\right)\)

PTHH: Cu(OH)₂ --t^o--> CuO + H₂O

_____0,05<-----------0,05<----0,05

=> m_Cu(OH)2 = (0,15-0,05).98 = 9,8 (g)

=> m_CuO = 0,05.80 = 4(g)

c) \(n_{SO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

=> \(n_{SO_2\left(pư\right)}=\dfrac{0,15.80}{100}=0,12\left(mol\right)\)

PTHH: Ca(OH)₂ + SO₂ --> CaSO₃\(\downarrow\) + H₂O

_____________0,12------>0,12

=> m_CaSO3 = 0,12.120 = 14,4(g)

a, \(n_{OH^-}=10^{-1}.V_A\left(mol\right)\)

\(n_{H^+}=10^{-2}.V_B\left(mol\right)\)

\(n_{OH^-dư}=10^{-2}.\left(V_A+V_B\right)\left(mol\right)\)

Ta có: \(n_{OH^-}-n_{OH^-dư}=n_{H^+}\)

\(\Leftrightarrow10^{-1}.V_A-10^{-2}.\left(V_A+V_B\right)=10^{-2}.V_B\)

\(\Leftrightarrow0,09V_A=0,02V_B\)

\(\Rightarrow\dfrac{V_A}{V_B}=\dfrac{2}{9}\)

b, Ta có: \(\left\{{}\begin{matrix}V_A+V_B=0,55\\\dfrac{V_A}{V_B}=\dfrac{2}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}V_A=0,1\left(l\right)\\V_B=0,45\left(l\right)\end{matrix}\right.\)

\(n_{BaCl_2}=\dfrac{1}{2}n_{Cl^-}=\dfrac{1}{2}n_{H^+}=\dfrac{1}{2}.10^{-2}.0,1=0,0005\left(mol\right)\Rightarrow m_{BaCl_2}=0,104\left(g\right)\)

\(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{OH^-dư}=\dfrac{1}{2}.10^{-2}.0,55=0,00275\left(mol\right)\Rightarrow m_{Ba\left(OH\right)_2}=0,047025\left(g\right)\)

\(\Rightarrow\%m_{Ba\left(OH\right)_2}=\dfrac{0,047025}{0,047025+0,104}.100\%=31,14\%\)

\(\Rightarrow\%m_{BaCl_2}=62,86\%\)

tham khảo 

\(C_{MddHCl}=\dfrac{n}{v}=>^nHCl=0,06\left(mol\right)\)

\(C_{MddBa\left(OH\right)_2}=\dfrac{n}{v}=>^nBa\left(OH\right)_2=0,04\left(mol\right)\)

\(2HCl+Ba\left(OH\right)_2->BaCl_2+2H_2O\)

0,06                              0,03

\(\dfrac{0,06}{3}< \dfrac{0,04}{1}=>Ba\left(OH\right)_2\)dư => làm quỳ tím chuyển màu xanh 

Khi cô cặn dd X thì có 0,01 mol \(Ba\left(OH\right)_2\)dư và 0,03 mol \(BaCl_2\)

=> \(^mcr=0,01.171+0,03.208=7,95\left(g\right)\)

Ta có : \(n_{H^+}=0,3\times0,05=0,015\left(mol\right)\)

\(n_{OH^-}=2\times a\times0,2=0,4a\left(mol\right)\)

Theo bài \(PH=12\) là môi trường bazo. Vậy OH- hết

PT: \(H^++OH^-\rightarrow H_2O\)

o,4a 0,4a (mol)

\(\Rightarrow H^+dư=0,015-0,4a\)

Ta có : \(\dfrac{0,015-0,4a}{0,5}=10^{-12}\Leftrightarrow a=0,375\left(M\right)\)

Ta có: \(\left\{{}\begin{matrix}n_{H^+}=0,015\left(mol\right)\\n_{OH^-}=0,2a\left(mol\right)\end{matrix}\right.\)

Ta có: pH = 12

=> \(\dfrac{0,015-0,2a}{0,5}=10^{-12}\)

\(\Rightarrow a=0,075\)

Đáp án D

pH =1 => [H+] = 0,1 => n H +  = 01.01 = 0,01 mol.

n O H -  = 0,1a

pH =12 => [H+] = 10^-12 => [OH-] =10^-2 => n O H -  dư = 0,01.(0,1+0,1) = 0,002 mol

=> 0,1a - 0,01 = 0,002 => a = 0,12.

\(n_{H^+}=0,2.0,3=0,06\left(mol\right)\)

\(n_{OH^-}=0,5.2.0,001V=0,001V\left(mol\right)\)

\(\Rightarrow n_{OH^-dư}=0,001V-0,06\left(mol\right)\)

\(\left[OH^-\right]=\dfrac{0,001V-0,06}{0,001V+0,3}=0,28\)

\(\Rightarrow V=200\left(ml\right)\)

\(n_{OH^-}=0,12V_1\)

\(n_{H^+}=0,17V_2\)

\(n_{OH^-dư}=\left(V_1+V_2\right).10^{-1}\)

Ta có: 

\(n_{OH^-dư}+n_{H^+}=n_{OH^-}\)

\(\Leftrightarrow\left(V_1+V_2\right).10^{-1}=0,12V_1\)

\(\Leftrightarrow0,1V_1=0,02V_2\)

\(\Rightarrow\dfrac{V_1}{V_2}=\dfrac{1}{5}\)