a) \(C_nH_{2n-6}+\dfrac{3n-3}{2}O_2\xrightarrow[]{t^o}nCO_2+\left(n-3\right)H_2O\)

b) \(C_xH_yO_z+\dfrac{2x+\dfrac{y}{2}-z}{2}O_2\xrightarrow[]{t^o}xCO_2+\dfrac{y}{2}H_2O\)

a) \(n_{HCl}=0,2.1,5=0,3\left(mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_{CuO}=x\left(mol\right)\\n_{Fe_2O_3}=y\left(mol\right)\end{matrix}\right.\) => 2x + y = 10 (*)

PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

              x----->2x

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

               y-------->6y

=> 2x + 6y = 0,3 (**)

Từ (*), (**) => \(\left\{{}\begin{matrix}x=0,075\\y=0,025\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,075.80}{10}.100\%=60\%\\\%m_{Fe_2O_3}=100\%-60\%=40\%\end{matrix}\right.\)

b) PTHH: \(CuO+CO\xrightarrow[]{t^o}Cu+CO_2\)

                 0,075->0,075

               \(Fe_2O_3+3CO\xrightarrow[]{t^o}2Fe+3CO_2\)

               0,025--->0,075

=> \(V_{CO_2}=\left(0,075+0,075\right).22,4=3,36\left(l\right)\)

Chỉ thu được muối axit => CO₂ dư, tính theo Ba(OH)₂ 

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

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

=> m = 0,1.259 = 25,9 (g)

1) \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH: \(Mg+H_2SO_{4\left(l\right)}\rightarrow MgSO_4+H_2\)

            \(Zn+H_2SO_{4\left(l\right)}\rightarrow ZnSO_4+H_2\)

            \(Fe+H_2SO_{4\left(l\right)}\rightarrow FeSO_4+H_2\)

Theo PT: \(n_{SO_4}=n_{H_2}=0,6\left(mol\right)\)

=> \(m=m_{muối}=m_{kl}+m_{SO_4}=33,1+0,6.96=90,7\left(g\right)\)

2) \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

            \(Zn+H_2SO_{4\left(l\right)}\rightarrow ZnSO_4+H_2\)

             \(Fe+H_2SO_{4\left(l\right)}\rightarrow FeSO_4+H_2\)

Theo PT: \(n_{SO_4}=n_{H_2}=0,6\left(mol\right)\)

=> \(m=m_{kl}=m_{muối}-m_{SO_4}=60,8-0,6.96=3,2\left(g\right)\)

3) \(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

Gọi hoá trị của R là n

PTHH: \(2R+2nHCl\rightarrow2RCl_n+nH_2\)

           \(\dfrac{0,9}{n}\)<--------------------------0,45

=> \(M_R=\dfrac{25,2}{\dfrac{0,9}{n}}=28n\left(g/mol\right)\)

Xét n = 2 thoả mãn =< M_R = 56 (g/mol)

=> R là Fe

4) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

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

          0,15<-----------------------0,15

=> \(m_{Fe}=0,15.56=8,4\left(g\right)\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{8,4}{10}.100\%=84\%\\\%m_{Cu}=100\%-84\%=16\%\end{matrix}\right.\)

`A + 2HCl -> ACl_2 + H_2`

`0,2`                              `0,2`       `(mol)`

`n_[H_2]=[4,48]/[22,4]=0,2(mol)`

`=>M_A=[11,2]/[0,2]=56(g//mol)`

  `=>A` là `Fe`

dòng thứ 2 mình chưa hiểu ý?

Độ tan của NaCl tại 20 ^oC là:

\(S^{20^{\circ}}=\dfrac{m_{ct}}{m_{dm}}\cdot100=\dfrac{7,18}{20}\cdot100=35,9\left(g\text{/}100ml\text{ }H_2O\right)\)

Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

nFe=0,1mol

PTHH: Fe+H2SO4=> FeSO4+H2

          0,1mol:0,2mol

          ta thấy nH2SO4 dư theo nFe

P/Ư:    0,1mol->0,1mol->0,1mol->0,1mol

=> thể tích H2 thu được sau phản ứng v=0,1.22,4=2,24ml

Ta có: \(n_{O_2}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\)

PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=0,15\left(mol\right)\)

\(\Rightarrow m_{KClO_3}=0,15.122,5=18,375\left(g\right)\)

\(\left(1\right)4FeS_2+11O_2\text{ }^{\underrightarrow{t^\circ}}8SO_2+2Fe_2O_3\)

\(\left(2\right)2SO_2+O_2\xrightarrow[^{V_2O_5}]{t^{\circ}}2SO_3\)

\(\left(3\right)SO_3+H_2O\rightarrow H_2SO_4\)

Câu (1) là SO4 không phải SO2 ạ