`1)`

`Y:Na:x(mol);Al:y(mol);Fe:z(mol)`

`->23x+27y+56z=2,16(1)`

Do nước dư nên NaOH hết hay Al dư

`2Na+2H_2O->2NaOH+H_2`

`2Al+2NaOH+2H_2O->2NaAlO_2+3H_2`

`->n_{H_2}=0,5x+1,5x={0,448}/{22,4}=0,02`

`->x=0,01(2)`

`n_{Al\ pu}=n_{NaOH}=x(mol)`

`->n_{Al\ du}=y-x(mol)`

`2Al+3CuSO_4->Al_2(SO_4)_3+3Cu`

`Fe+CuSO_4->FeSO_4+Cu`

`->1,5(y-x)+z=n_{Cu}={3,2}/{64}=0,05(3)`

`(1)(2)(3)->x=0,01;y=0,03;z=0,02`

`m_{Na}=0,01.23=0,23(g)`

`m_{Al}=0,03.27=0,81(g)`

`m_{Fe}=0,02.56=1,12(g)`

`2)`

Giả sử `m_A=110(g)` 

`->m_{Al_2O_3}={110}/{1+10}=10(g)`

`A:CaCO_3:x(mol);MgCO_3:y(mol)`

`->100x+84y=110-10=101(1)`

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

`MgCO_3` $\xrightarrow{t^o}$ `MgO+CO_2`

`->n_{CaO}=x(mol);n_{MgO}=y(mol)`

`->56x+40y+10=56,8\%.110=62,48(2)`

`(1)(2)->x={1151}/{2200};y={51}/{88}`

`\%m_{Al_2O_3}={10}/{110}.100\%=9,09\%`

`\%m_{CaCO_3}={100x}/{110}.100\%=47,56\%`

`\%m_{MgCO_3}=100-9,09-47,56=43,35\%`