Đáp án:

`\%Mg=75,68\%`

`\%Al=24,32\%`

Giải thích các bước giải:

Trong X: `n_{Cl_2}+n_{O_2}={8,6765}/{24,79}=0,35(1)`

Bảo toàn K ...

`CH_3-CHOH-CH_3` $\xrightarrow[H^+]{170^oC}$ `CH_2=CH-CH_3+H_2O`

`a)`

`n_{CaCO_3\ l t}={500.80\%}/{100}=4(mol)`

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

Theo PT: `n_{CO_2\ l t}=n_{CaCO_3\ l t}=4(mol)`

Bảo toàn KL: `m_X=500-4.70\ ...

`a)`

`C_6^0->6C^{+4}+24e|xx1`

`O_2^0+4e->2O^{-2}|xx6`

`C_6H_{12}O_6+6O_2` $\xrightarrow{t^o}$ `6CO_2+6H_2O`

`b)`

`n_{C_6H_{12}O_6}={10000}/{2800}={25}/7(mol ...

`a)`

`C_3^{-8//3}->3C^{+4}+20e|xx1`

`O_2^0+4e->2O^{-2}|xx5`

`C_3H_8+5O_2` $\xrightarrow{t^o}$ `3CO_2+4H_2O`

`C_4^{-5//2}->4C^{+4}+26e|xx2`

`O_2^0+4e-> ...

`a)`

`n_{Fe}={16,8}/{56}=0,3(mol)`

`Fe+2HCl->FeCl_2+H_2`

`0,3->0,6->0,3->0,3(mol)`

`->V_{H_2}=0,3.24,79=7,437(L)`

`b)` `m_{dd\ HCl}={0,6.36,5}/{14, ...

`CH_2=CH_2+Br_2->CH_2Br-CH_2Br`

`CH_2Br-CH_2Br+2KOH->CH≡CH+2KBr+2H_2O`

`CH≡CH+2[Ag(NH_3)_2]OH->CAg≡CAg+4NH_3+2H_2O`

`CAg≡CAg+2HCl->2AgCl+CH& ...

`C_4H_{10}` $\xrightarrow{t^o,xt,p}$ `C_3H_6+CH_4`

`nCH_2=CHCH_3` $\xrightarrow{t^o,p,xt}$ `(-CH_2-CH(CH_3)-)_n`

`3CH_2=CHCH_3+2KMnO_4+4H_2O->3CH_2OH-CHOH-CH_3+2MnO_2+2KOH`

...

`HC≡CH+HCl` $\xrightarrow{t^o}$ `CH_2=CHCl`

`nCH_2=CHCl` $\xrightarrow{t^o,p,xt}$ `(-CH_2-CH(Cl)-)_n`

`Fe_3^{+8//3}->3Fe^{+3}+1e|xx3`

`N^{+5}+3e->N^{+2}|xx1`

`3Fe_3O_4+28HNO_3->9Fe(NO_3)_3+NO+14H_2O`

`FeS_2->Fe^{+3}+2S^{+6}+15e|xx1`

`N^{+5}+1e->N^{+4}|xx15 ...