Giải thích các bước giải:

nH2S = 3,36/22,4= 0,15 mol 

- Giả sử muối sinh ra là Na2S

2NaOH + H2S → Na2S + 2H2O

                0,15      0,15              (mol)

Theo PT ⇒ nNa2S = nH2S = 0,15 mol

⇒ m muối = 0,15.78 = 11,7 gam

\(n_{NaOH}=0,25.1=0,25\left(mol\right)\\ n_{H_2S}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(T=\dfrac{0,25}{0,15}=\dfrac{5}{3}\rightarrow\) Tạo cả 2 muối

Gọi \(\left\{{}\begin{matrix}n_{NaOH\left(\text{tạo muối trung hoà}\right)}=a\left(mol\right)\\n_{NaOH\left(\text{tạo muối axit}\right)}=b\left(mol\right)\end{matrix}\right.\)

PTHH:

2NaOH + H₂S ---> Na₂S + 2H₂O

a                \(\dfrac{a}{2}\)          \(\dfrac{a}{2}\)

NaOH + H₂S ---> NaHS + H₂O

b              b             b

Hệ pt \(\left\{{}\begin{matrix}a+b=0,25\\\dfrac{a}{2}+b=0,15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,05\left(mol\right)\end{matrix}\right.\)

\(\rightarrow m_{muối}=\dfrac{0,2}{2}.78+0,05.56=10,6\left(g\right)\)

Theo giả thiết ta có: \(n_{Al_2\left(SO_4\right)_3}=0,08\left(mol\right)\)

\(Al_2\left(SO_4\right)_3+3Ba\left(OH\right)_2\rightarrow2Al\left(OH\right)_3+3BaSO_4\) (1)

\(2Al\left(OH\right)_3+Ba\left(OH\right)_2\rightarrow Ba\left(AlO_2\right)_2+4H_2O\) (2)

Giả sử kết tủa chỉ là BaSO₄ thì nhận thấy \(n_{\downarrow}< n_{SO_4}\). Do đó  chỉ xảy ra phản ứng (1).

Gọi số mol Ba(OH)₂ là x (mol)

Bảo toàn nguyên tố ta có: \(n_{BaSO_4}=x\left(mol\right);n_{Al\left(OH\right)_3}=\dfrac{2x}{3}\left(mol\right)\)

\(\Rightarrow233x+78.\dfrac{2x}{3}=51,3\Rightarrow x=0,18\)

\(\Rightarrow V=\dfrac{0,18}{2}=0,09\left(l\right)\)

\(\dfrac{n_{OH^{^{ }-}}}{n_{CO_2}}=\dfrac{0,125.2}{\dfrac{3,36}{22,4}}=1,67\Rightarrow Tạo.2.muối\\ Ba\left(OH\right)_2+CO_2->BaCO_3+H_2O\\ BaCO_3+CO_2+H_2O->Ba\left(HCO_3\right)_2\\ n_{BaCO_3max}=0,125mol=n_{CO_2\left(1\right)}\\ n_{Ba\left(HCO_3\right)_2}=0,15-0,125=0,025mol\\ C_{M\left(Ba\left(HCO_3\right)_2\right)}=\dfrac{0,025}{0,125}=0,2M\)

Tham khảo:

nCO2=0,15 mol, nBa(OH)2=0,125 mol

1<nOH-/nCO2=0,25/0,15=1,67<2 => Tạo 2 muối

BaCO3: x

Ba(HCO3)2: y

x+y=nBa2+=0,125

x+2y=nC=0,15

=>x=0,1; y=0,025

CM Ba(HCO3)2=0,025/0,125=0,2M

\(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(n_{Ba\left(OH\right)_2}=0,125\cdot1=0,125\left(mol\right)\)

\(T=\dfrac{n_{Ba\left(OH\right)_2}}{n_{CO_2}}=\dfrac{0,125}{0,15}=\dfrac{5}{6}\Rightarrow\dfrac{1}{2}< T< 1\)

⇒ Phản ứng tạo ra 2 muối \(Ba\left(HCO_3\right)_2\) và \(BaCO_3\downarrow\).

Gọi số mol \(\left\{{}\begin{matrix}Ba\left(HCO_3\right)_2:x\\BaCO_3:y\end{matrix}\right.\left(mol\right).ĐK:x;y>0\)

PTPƯ:

\(Ba\left(OH\right)_2+2CO_2\rightarrow Ba\left(HCO_3\right)_2\)

       x         ←    2x    ←          x

\(Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3\downarrow+H_2O\)

       y       ←     y     ←    y

Ta có hpt:

\(\left\{{}\begin{matrix}2x+y=0,15\\x+y=0,125\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,025\\y=0,1\end{matrix}\right.\)

Nồng độ chất tan:

\(C_M\left[\left(BaHCO_3\right)_2\right]=\dfrac{0,025}{0,125}=0,2\left(M\right)\)

\(n_{BaCO_3}=\dfrac{19.7}{197}=0.1\left(mol\right)\)

\(n_{Ba\left(OH\right)_2}=0.15\cdot1=0.15\left(mol\right)\)

\(n_{MgCO_3}=a\left(mol\right),n_{CaCO_3}=b\left(mol\right)\)

\(\Rightarrow m_A=84a+100b=18.4\left(g\right)\left(1\right)\)

\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)

\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

\(n_{CO_2}=a+b\left(mol\right)\)

TH1 : Không tạo muối axit , Ba(OH)₂ dư 

\(\Rightarrow n_{CO_2}=n_{BaCO_3}=0.1\left(mol\right)\)

\(\Rightarrow a+b=0.1\left(2\right)\)

\(\left(1\right),\left(2\right):a=-0.525,b=0.625\left(L\right)\)

TH2 : Phản ứng tạo hai muối vừa đủ 

\(n_{CO_2}=0.1+\left(0.15-0.1\right)\cdot2=0.2\left(mol\right)\)

\(\Rightarrow a+b=0.1\left(3\right)\)

\(\left(1\right),\left(3\right):a=b=0.1\)

\(\%MgCO_3=\dfrac{8.4}{18.4}\cdot100\%=45.65\%\)

\(\%CaCO_3=54.35\%\)

giúp mình với .mình ko bik lm

`n_{Ba(OH)_2}=0,135.2=0,27(mol)`

`n_{Al_2(SO_4)_3}=0,08.1=0,08(mol)`

`Al_2(SO_4)_3+3Ba(OH)_2->2Al(OH)_3+3BaSO_4`

`0,08->0,24->0,16->0,24(mol)`

`->n_{Ba(OH)_2\ du}=0,27-0,24=0,03(mol)`

`Ba(OH)_2+2Al(OH)_3->Ba(AlO_2)_2+4H_2O`

`0,03->0,06(mol)`

`->n_{Al(OH)_3\ du}=0,16-0,06=0,1(mol)`

`2Al(OH)_3`  $\xrightarrow{t^o}$  `Al_2O_3+3H_2O`

`0,1->0,05(mol)`

`->a=m_{Al_2O_3}+m_{BaSO_4}=0,05.102+0,24.233=61,02(g)`

a, \(n_{H2S}=0,1\left(mol\right)\)

\(n_{NaOH}=0,1.2=0,2\left(mol\right)\)

\(\frac{n_{NaOH}}{n_{H2S}}=2\)

\(2NaOH+H_2S\rightarrow Na_2S+2H_2O\)

\(n_{Na2S}=n_{H2S}=0,1\left(mol\right)\)

\(\Rightarrow m_{Na2S}=7,8\left(g\right)\)

b, \(n_{KOH}=0,1\left(mol\right)\)

\(\frac{n_{KOH}}{n_{H2S}}=1\)

\(KOH+H_2S\rightarrow KHS+H_2O\)

\(\Rightarrow n_{KHS}=n_{H2S}=0,1\left(mol\right)\)

\(\Rightarrow m_{KHS}=7,2\left(g\right)\)

c, \(n_{NaOH}=0,12\left(mol\right)\)

\(\frac{n_{NaOH}}{n_{H2S}}=1,2\)

\(2NaOH+H_2S\rightarrow Na_2S+2H_2O\)

\(NaOH+H_2S\rightarrow NaHS+H_2O\)

\(\Rightarrow\left\{{}\begin{matrix}2a+b=0,12\\a+b=0,1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,02\\b=0,08\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Na2S}=0,02.78=1,56\left(g\right)\\m_{NaHS}=0,08.56=4,48\left(g\right)\end{matrix}\right.\)