Ta có: \(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

PTHH: \(2M+xH_2SO_4\rightarrow M_2\left(SO_4\right)_x+xH_2\)

           0,02--------------------------------->0,01x

`=>` \(0,01x=0,03\Leftrightarrow x=3\)

Vậy M có hoá trị III

Giả sử có 1 mol M phản ứng với dd CuSO₄

PTHH: \(2M+3CuSO_4\rightarrow M_2\left(SO_4\right)_3+3Cu\downarrow\)

              1-------------------------------------->1,5

`=>` \(\dfrac{1,5.64}{M_M}=3,555\Leftrightarrow M_M=27\left(g/mol\right)\)

`=> M` là `Al` (nhôm)

Giả sử có 1 mol Fe phản ứng

PTHH: \(2Fe+6H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)

             1---->3---------------->0,5------------->1,5

`=>` \(m_{Fe_2\left(SO_4\right)_3}=0,5.400=200\left(g\right)\)

Ta có: \(C\%_{Fe_2\left(SO_4\right)_3}=C\%_{H_2SO_4\left(dư\right)}\)

`=>` \(m_{H_2SO_4\left(dư\right)}=m_{Fe_2\left(SO_4\right)_3}=200\left(g\right)\)

`=>` \(n_{H_2SO_4\left(dư\right)}=\dfrac{200}{98}=\dfrac{100}{49}\left(mol\right)\)

`=>` \(m_{ddH_2SO_4}=\dfrac{\left(\dfrac{100}{49}+3\right).98}{78,4\%}=\dfrac{30875}{49}\left(g\right)\)

`=>` \(m_{ddspư}=\dfrac{30875}{49}+56-1,5.64=\dfrac{28915}{49}\left(g\right)\)

`=>` \(C\%_{Fe_2\left(SO_4\right)_3}=C\%_{H_2SO_{4\left(dư\right)}}=\dfrac{200}{\dfrac{28915}{49}}.100\%=33,7\%\)

`Fe + 2HCl -> FeCl_2 + H_2 \uparrow`

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

`n_[H_2]=[3,7185]/[24,79]=0,15(mol)`

`a)%m_[Fe]=[0,15.56]/18 .100=46,67%`

`=>%m_[Cu]=100-46,67=53,33%`

`b)C_[M_[FeCl_2]]=[0,15]/[0,1]=1,5(M)`

`c)Cu + 2H_2 SO_[4(đ,n)] -> CuSO_4 + SO_2\uparrow + 2H_2 O`

 `0,125`         `0,25`                     `0,125`                                          `(mol)`

`n_[Cu]=[18-0,15.56]/64=0,15(mol)`

`n_[H_2 SO_[4(đ,n)]]=[25.98]/[100.98]=0,25(mol)`

Có: `[0,15] > [0,25]/2->Cu` dư

  `=>m_[CuSO_4]=0,125.160=20(g)`

\(n_{CO_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\Rightarrow n_C=0.2\left(mol\right)\)

\(n_{H_2O}=\dfrac{7.2}{18}=0.4\left(mol\right)\Rightarrow n_H=0.4\cdot2=0.8\left(mol\right)\)

\(n_C:n_H=0.2:0.8=1:4\)

CT của A : CH₄

\(CH_4+2O_2\underrightarrow{^{t^0}}CO_2+2H_2O\)

\(CH_4+Cl_2\underrightarrow{^{^{as}}}CH_3Cl+HCl\)

\(CH_4+2Cl_2\underrightarrow{^{^{as}}}CH_2Cl_2+2HCl\)

\(CH_4+2Cl_2\underrightarrow{^{^{as}}}CHCl_3+HCl\)

Dẫn lần lượt các khí vào bình đựng Ca(OH)₂ dư : 

- Kết tủa trắng : CO₂

Cho tàn que đóm đỏ lần lượt vào lọ khí còn lại : 

- Bùng cháy : O₂

- Khí cháy với ngọn lửa xanh nhạt : H₂

- Tắt hẳn : CH₄

\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)

\(KCl+AgNO_3\rightarrow KNO_3+AgCl\downarrow\\ ZnCl_2+2AgNO_3\rightarrow Zn\left(NO_3\right)_2+2AgCl\downarrow\\ Đặt:n_{KCl}=a\left(mol\right);n_{ZnCl_2}=b\left(mol\right)\left(a,b>0\right)\\ \Leftrightarrow\left\{{}\begin{matrix}74,5a+136b=5,7\\143,5a+287b=11,48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,04\\b=0,02\end{matrix}\right.\\ a,\%m_{KCl}=\dfrac{74,5.0,04}{5,7}.100\approx52,281\%\\ \%m_{ZnCl_2}=\dfrac{0,02.136}{5,7}.100\approx47,719\%\\ b,n_{NaOH}=0,025.2=0,05\left(mol\right)\\ Zn\left(NO_3\right)_2+2NaOH\rightarrow Zn\left(OH\right)_2+2NaNO_3\\ Vì:\dfrac{0,02}{1}< \dfrac{0,05}{2}\Rightarrow NaOH\left(dư\right)\\ n_{NaOH\left(Dư\right)}=0,05-0,02.2=0,01\left(mol\right)\\ 2NaOH_{dư}+Zn\left(OH\right)_2\rightarrow Na_2ZnO_2+2H_2O\)

\(n_{Zn\left(OH\right)_2\left(p.ứ\right)}=\dfrac{0,01}{2}=0,005\left(mol\right)\\ m_{Zn\left(OH\right)_2\left(còn\right)}=m_{kt}=\left(0,02-0,005\right).99=1,485\left(g\right)\)

a) Gọi \(\left\{{}\begin{matrix}n_{Fe}=x\left(mol\right)\\n_{Fe_2O_3}=y\left(mol\right)\end{matrix}\right.\)

`=> 56x + 160y = 13,6 (1)`

Ta có: \(n_{HCl}=\dfrac{91,25.20\%}{36,5}=0,5\left(mol\right)\)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

            x---->2x-------->x-------->x

            \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

            y--------->6y--------->2y

`=> 2x + 6y = 0,5 (2)`

`(1), (2) => x = 0,1; y = 0,05`

`=>` \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{13,6}.100\%=41,18\%\\\%m_{Fe_2O_3}=100\%-41,18\%=58,82\%\end{matrix}\right.\)

b) \(m_{ddspư}=91,25+13,6-0,1.2=104,65\left(g\right)\)

`=>` \(\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,1.127}{104,65}.100\%=12,14\%\\C\%_{FeCl_3}=\dfrac{0,05.2.162,5}{104,65}.100\%=15,53\%\end{matrix}\right.\)

c) Ta có: \(n_{NaOH}=\dfrac{64.1,25.10\%}{40}=0,2\left(mol\right)\)

PTHH: \(2Fe+6H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}Fe_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)

            0,1----------------------------------------->0,15

            \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

Xét \(T=\dfrac{n_{NaOH}}{n_{SO_2}}=\dfrac{0,2}{0,15}=\dfrac{4}{3}\)

`=>` Pư tạo 2 muối

PTHH: \(2NaOH+SO_2\rightarrow Na_2SO_3+H_2O\)

             0,2--------->0,1----->0,1

            \(Na_2SO_3+SO_2+H_2O\rightarrow2NaHSO_3\)

             0,05<-----0,05--------------->0,1

`=>` \(\left\{{}\begin{matrix}C_{M\left(Na_2SO_3\right)}=\dfrac{0,1-0,05}{0,064}=0,78125M\\C_{M\left(NaHSO_3\right)}=\dfrac{0,1}{0,064}=1,5625M\end{matrix}\right.\)

Câu 89:

Câu 95:

A: Fe(OH)₃

B: FeCl₃

C: Fe₂O₃

(1) \(2Fe+6H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}Fe_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)

(2) \(Fe_2\left(SO_4\right)_3+3Ba\left(OH\right)_2\rightarrow2Fe\left(OH\right)_3\downarrow+3BaSO_4\downarrow\)

(3) \(2Fe\left(OH\right)_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+6H_2O\) 

(5) \(2Fe+3Cl_2\xrightarrow[]{t^o}2FeCl_3\)

(6) \(FeCl_3+3KOH\rightarrow Fe\left(OH\right)_3\downarrow+3KCl\)

(7) \(2Fe\left(OH\right)_3\xrightarrow[]{t^o}Fe_2O_3+3H_2O\)

(8) \(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+3H_2O\)

Câu 96:

`A: SO_2`

`B: Fe_2O_3`

`C: SO_3`

`E: H_2SO_4`

`F: CuSO_4`

`D: H_2O`

`4FeS_2 + 11O_2 -> (t^o) 2Fe_2O_3 + 8SO_2`

`2SO_2 + O_2 -> (t^o, V_2O_5) 2SO_3`

`SO_3 + H_2O -> H_2SO_4`

`2H_2SO_4 + Cu -> CuSO_4 + SO_2 + 2H_2O`

`SO_2 + H_2O -> H_2SO_3`