`a)`

`2Al+6HCl->2AlCl_3+3H_2`

`Fe+2HCl->FeCl_2+H_2`

`Zn+2HCl->ZnCl_2+H_2`

`b)`

Áp dụng định luật bảo toàn khối lượng, ta có:

\(m_{muối}=m_{hh}+m_{HCl}-m_{H_2}\)

          \(=29,6+51,1-1,4=79,3\left(g\right)\)

`1)`\(2Ca+O_2\rightarrow\left(t^o\right)2CaO\)

`2)CaO+2HCl->CaCl_2+H_2O`

`3)3Fe+2O_2->(t^o)Fe_3O_4`

`4)Fe_3O_4+4H_2->3Fe+4H_2O`

`5)2Al+3H_2SO_4->Al_2(SO_4)_3+3H_2O`

`6)3Mg+Al_2(SO_4)_3->3MgSO_4+2Al`

`7)MgSO_4+2KOH->Mg(OH)_2+K_2SO_4`

`8)Mg(OH)_2+2HNO_3->Mg(NO_3)_2+2H_2O`

`9)Fe(OH)_3+3HCl->FeCl_3+3H_2O`

`10)2Fe(OH)_3+3H_2SO_4->Fe_2(SO_4)_3+6H_2O`

\(n_{HCl}=\dfrac{18,25.8\%}{36,5}=0,04\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,02     0,04         0,02              ( mol )

\(m_{Zn}=0,02.65=1,3\left(g\right)\)

\(m_{ZnCl_2}=0,02.136=2,72\left(g\right)\)

`a)`\(Fe_xO_y+yH_2\rightarrow\left(t^o\right)xFe+yH_2O\)

`b)`\(Fe_xO_y+2yHCl\rightarrow xFeCl_{\dfrac{2y}{x}}+yH_2O\)

`c)`\(2Fe_xO_y+\left(\dfrac{3x-2y}{2}\right)O_2\rightarrow\left(t^o\right)xFe_2O_3\)

`d)`\(C_xH_y+\left(x+\dfrac{y}{4}\right)O_2\rightarrow\left(t^o\right)xCO_2+\dfrac{y}{2}H_2O\)

`e)`\(C_xH_yO_z+\left(x+\dfrac{y}{4}-\dfrac{z}{2}\right)O_2\rightarrow\left(t^o\right)xCO_2+\dfrac{y}{2}H_2O\)

\(n_{H_2}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\)

Gọi kim loại đó là R

\(R+2H_2O\rightarrow R\left(OH\right)_2+H_2\)

0,125                                 0,125   ( mol )

\(M_R=\dfrac{m}{n}=\dfrac{17,125}{0,125}=137\) \((g/mol)\)

Vậy R là Bari ( Ba )

`b)`\(5x-3y=2xy-11\)

\(\Leftrightarrow5x-3y-2xy+11=0\)

\(\Leftrightarrow10x-6y-4xy+22=0\)

\(\Leftrightarrow2x\left(5-2y\right)+3\left(5-2y\right)+7=0\)

\(\Leftrightarrow\left(5-2y\right)\left(2x+3\right)=-7\)

Kẻ bảng giá trị là xong á....

`c)`\(x^2-2x-11=y^2\) ( sửa đề )

\(\Leftrightarrow\left(x^2-2x+1\right)-y^2-12=0\)

\(\Leftrightarrow\left(x-1\right)^2-y^2=12\)

\(\Leftrightarrow\left(x-1-y\right)\left(x-1+y\right)=12\)

Kẻ bảng giá trị....

`a)`\(11x+18y=120\)

\(\Leftrightarrow x=\dfrac{120-18y}{11}\)

\(\Leftrightarrow x=\dfrac{121-11y-7y-1}{11}\)

\(\Leftrightarrow x=11-y-\dfrac{y+1}{11}\)

Vì \(x;y\in Z\) nên để \(x;y\in Z\) thì \(\dfrac{y+1}{11}\in Z\)

Đặt \(\dfrac{y+1}{11}=k;k\in Z\)

\(\Leftrightarrow y+1=11k\)

\(\Leftrightarrow y=11k-1\)

\(\Rightarrow x=11-\left(11k-1\right)-k\)

\(\Leftrightarrow x=11-11k+1-k\)

\(\Leftrightarrow x=12-12k\)

Vậy nghiệm nguyên của pt: \(\left\{{}\begin{matrix}x=12-12k\\y=11k-1;k=\dfrac{y+1}{11}\in Z\end{matrix}\right.\)

`a)`\(\left(x-3256\right)+2589=12587-2658\)

\(\Rightarrow x-3256=12587-2658-2589\)

\(\Rightarrow x-3256=12587-5247\)

\(\Rightarrow x-3256=7340\)

\(\Rightarrow x=7340+3256\)

\(\Rightarrow x=10596\)

`b)`\(125879-\left(x-2597\right)-5879=52369\)

`=>`\(125879-\left(x-2597\right)=52369+5879\)

`=>`\(125879-\left(x-2597\right)=58248\)

`=>`\(x-2597=125879-58248\)

`=>`\(x-2597=67631\)

`=>`\(x=2597+67631\)

`=>x=70228`

`a)` Ta có: `AE////BC`

\(\Rightarrow\widehat{y}=\widehat{EAB}=70^o\) ( so le trong )

\(\widehat{x}+\widehat{y}+\widehat{ACB}=180^o\) ( tổng 3 góc )

\(\Rightarrow\widehat{x}=180^o-70^o-75^o=35^o\)

`b)`Ta có: `AE////BC`

\(\Rightarrow\widehat{y}=\widehat{EAC}=60^o\) ( so le trong )

\(\widehat{x}+\widehat{y}+\widehat{ABC}=180^o\) ( tổng 3 góc )

\(\Rightarrow\widehat{x}=180^o-60^o-90^o=30^o\)

Bài 3.

\(n_{CaO}=\dfrac{11,2}{56}=0,2\left(mol\right)\); \(n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)\)

\(CaCO_3\rightarrow\left(t^o\right)CaO+CO_2\)

   1                        1          1     ( mol )

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

\(m_{CaCO_3}=n.M=0,2.100=20\left(g\right)\)

Bài 4.
`a)`\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\left(2:1:2\right)\)

`b)`\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\left(3:2:1\right)\)

`c)`\(C_{12}H_{22}O_{11}\rightarrow\left(H_2SO_4\right)12C+11H_2O\left(1:12:11\right)\)

`d)`\(4FeO+O_2\rightarrow\left(t^o\right)2Fe_2O_3\left(4:1:2\right)\)

`e)`\(N_2O_5+H_2O\rightarrow2HNO_3\left(1:1:2\right)\)

`f)`\(2Fe\left(OH\right)_3\rightarrow\left(t^o\right)Fe_2O_3+3H_2O\left(2:1:3\right)\)

`g)`\(4Na+O_2\rightarrow\left(t^o\right)2Na_2O\left(4:1:2\right)\)

`h)`\(Na_2CO_3+CaCl_2\rightarrow CaCO_3+2NaCl\left(1:1:1:2\right)\)

`i)`\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\left(2:2:3\right)\)