\(9.1\\ n_{NaOH}=1,5.0,2=0,3\left(mol\right)\\ n_{H_3PO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\\ Vì:n_{H_3PO_4}:n_{NaOH}=0,3:0,3=1:1\\ \Rightarrow Sp:Duy.nhất.NaH_2PO_4\\ ChọnB\\ 9.2\\ NaOH+H_3PO_4\rightarrow NaH_2PO_4+H_2O\\ n_{muối}=n_{NaOH}=0,3\left(mol\right)\\ m_{muối}=0,3.120=36\left(g\right)\\ Chọn.A\)

Để thu được duy nhất muối K₂HPO₄ thì tỉ lệ kiềm: axit = 2:1

\(10.1\\ n_{KOH}=0,3.0,6=0,18\left(mol\right)\\ n_{H_3PO_4}=0,18:2=0,09\left(mol\right)\\ V=V_{ddH_3PO_4}=\dfrac{0,09}{0,3}=0,3\left(l\right)\\ Chọn.C\\ 10.2\\ 2KOH+H_3PO_4\rightarrow K_2HPO_4+2H_2O\\ n_{muối}=n_{H_3PO_4}=0,09\left(mol\right)\\ m_{muối}=0,09.174=15,66\left(g\right)\\ Chọn.C\)

tỉ lệ mol H₃PO₄:Ca(OH)₂ = 1:1 --> Tạo muối CaHPO₄ --> m muối= 20,4g 

`14)`

`H_3 PO_4 + 3KOH -> K_3 PO_4 + 3H_2 O`

    `0,05`      `0,15`                                           `(mol)`

`n_[H_3 PO_4]=1.0,05=0,05(mol)`

  `=>V_[dd KOH]=[0,15]/1=0,15(l)=150(ml)->\bb A`

___________________________________________

`15)`

`n_[OH^-]=0,45.2.0,5=0,45(mol)`

`n_[PO_4 ^[3-]]=2.0,075=0,15(mol)`

  `=>[n_[OH^-]]/[n_[PO_4 ^[3-]]]=3=>` Tạo muối `PO_4 ^[3-]`

`2H_3 PO_4 + 3Ca(OH)_2 -> Ca_3 (PO_4)_2 + 6H_2 O`

    `0,15`            `0,225`                  `0,075`                              `(mol)`

  `=>m_[Ca_3 (PO_4)_2]=0,075.310=23,25(g)->\bb A`

`

`n_[CaCO_3]=[10^6 . 90/100]/100=9000 (mol)`

`CaCO_3` $\xrightarrow{t^o}$ `CaO + CO_2 \uparrow`

 `9000`               `9000`   `9000`         `(mol)`

`V_[CO_2]=9000.22,4=201600(l)`

10^6 là gì vậy ạ

a) Gọi \(\left\{{}\begin{matrix}n_{H_2}=a\left(mol\right)\\n_{CO_2}=b\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2a+44b=4,8\\a+b=\dfrac{6,72}{22,4}=0,3\end{matrix}\right.\)

=> a = 0,2; b = 0,1

=> \(\left\{{}\begin{matrix}\left\{{}\begin{matrix}\%V_{H_2}=\dfrac{0,2}{0,3}.100\%=66,67\%\\\%V_{CO_2}=100\%-66,67\%=33,33\%\end{matrix}\right.\\\left\{{}\begin{matrix}\%m_{H_2}=\dfrac{0,2.2}{4,8}.100\%=8,33\%\\\%m_{CO_2}=100\%-8,33\%=91,67\%\end{matrix}\right.\end{matrix}\right.\)

b) \(M_{TB}=\dfrac{4,8}{0,3}=16\left(g/mol\right)\)

\(2C_4H_{10}+13O_2->8CO_2+10H_2O\)

\(2C_4H_{10}+13O_2\overset{t^o}{--->}8CO_2+10H_2O\)