B1

a)1:2:1:1

b)nMg=0,396

theo pt:  nH2=nMg=0,396...

->mH2=nH2.MH2=0,396.2=0,791

MÌNH LÀM TRÒN SỐ NHA

b2

4:3:2    (Fe2O3)

1:2:1:1

a) PTHH

`Mg` + `HCl` $\dashrightarrow$ `MgCl_2` + `H_2` `\uparrow`

`Mg` + `2HCl` `->` `MgCl_2` + `H_2` `\uparrow`

b) 

`n_{MgCl_2}` = `m/M` = `9.5/95` = `0.1  mol`

`n_{MgCl_2}` = `n_{H_2}` = `0.1  mol`

`=>` `m_{H_2}` = `n.M` = `0.1 . 2` = `0.2  mol` 

3.

a) `Fe` + `O_2` $\dashrightarrow$ `Fe_3 O_4` 

    `Fe` + `O_2` $\xrightarrow{t^o}$ `Fe_3 O_4` 

b)  `CaO + HCl` $\dashrightarrow$ `CaCl_3 + H_2 O`

     `CaO + HCl` `->` `CaCl_3 + H_2 O`

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