`n_{CaCO_3}=10/100=0,1(mol)`

`PTHH:2CH_3COOH+Mg ->(CH_3COO)_2Mg+H_2`

           `2CH_3COOH+MgCO_3->(CH_3COO)_2Mg+CO_2+H_2O`

            `CO_2+Ca(OH)_2->CaCO_3+H_2O`

`a.n_{MgCO_3}=n_{CO_2}=n_{CaCO_3}=0,1(mol)`

`%MgCO_3={0,1.84.100%}/{14,4}~~58,3%`

`%Mg=100-58,3=41,7%`

hh B: $\rm H_2,CO_2$

`n_{Mg}={14,4-0,1.84}/24=0,25(mol)`

`n_{H_2}=n_{Mg}=0,25(mol)`

`%H_2={0,25}/{0,25+0,1} .100%~~71,4%`

`%CO_2=100-71,4=28,6%`

`b.`

`n_{CH_3COOH}=2n_{Mg}=2n_{MgCO_3}=0,25.2+0,1.2=0,7(mol)`

`V_{CH_3COOH}={0,7}/2=0,35(l)=350(ml)`

`m_{dd CH_3COOH}=350.1,2=420(g)`

$\rm n_{CaCO_3}=\dfrac{10}{100}=0,1(mol)$

  $\rm pthh : Mg + 2CH_3COOH \xrightarrow{} (CH_3COO)_2Mg +H_2$  (1)    

      $\rm MgCO_3 +2CH_3COOH \xrightarrow{} (CH_3COO)_2Mg +CO_2 +H_2O$  (2)

          $\rm CO_2 +Ca(OH)_2 \xrightarrow{} CaCO_3 +H_2O$  (3)

theo pthh (2) 

$\rm n_{MgCO_3}=n_{CO_2}=0,1(mol)$ 

`->`$\rm m_{MgCO_3}=0,1.84=8,4(g)$ 

`->`$m_{Mg}=14,4-8,4=6(g)$ 

`=>`$n_{Mg}=\dfrac{6}{24}=0,25(mol)$ 

theo pthh (1)  $n_{H_2}=n_{Mg}=0,25(mol)$ 

$\%V_{H_2}=\dfrac{0,25}{0,25+0,1}.100\%=71,4\%$ 

`->`$\%V_{CO_2}=100\%-71,4\%=28,6\%$

`b)` 

theo pthh (1),(2) 

$n_{CH_3COOH}=2.(0,25+0,1)=0,7(mol)$ 

ta có $C_{M_{CH_3COOH}}=\dfrac{n}{V}=2M$ 

     `<=>`$\dfrac{0,7}{V}=2$ 

     `<=>`$V=0,35(l)=350(ml)$ 

`->`$m_{CH_3COOH}=350.1,2=420(g)$