\(\begin{array}{l}
a)\\
Al_2O_3+2NaOH\to 2NaAlO_2+H_2O\\
\to m_{MgO}=40(g)\\
\to \%m_{MgO}=\frac{40}{50,2}.100\%≈79,7\%\\
\to \%m_{Al_2O_3}=100-79,7=20,3\%\\
b)\\
200ml=0,2l\\
n_{Al_2O_3}=\frac{50,2-40}{102}=0,1(mol)\\
Theo\,PT:\,n_{NaOH}=2n_{Al_2O_3}=0,2(mol)\\
\to C_{M\,NaOH}=\frac{0,2}{0,2}=1M\\
c)\\
m_{dd\,NaOH}=200.1,15=230(g)\\
m_{dd\,spu}=50,2+230-40=240,2(g)\\
Theo\,PT:\,n_{NaAlO_2}=n_{NaOH}=0,2(mol)\\
\to C\%_{NaAlO_2}=\frac{0,2.82}{240,2}.100\%\approx 6,83\%
\end{array}\)

`#Ly`

$a)$

PTHH: $Al_2O_3$ $+$ $2NaOH$ `->` $2NaAlO_2$ $+$ $H_2O$

$($ $MgO$ không pư với $NaOH)$

`->` Chất rắn không tan thu được: $MgO$

`%m_(MgO` `=` `(40)/(50,2).100%≈79,68%`

`%m_(Al_2O_3` `=` `100-79,68=20,32%`

$b)$

`m_(Al_2O_3` `=` `50,2-40=10,2` $(g)$

`n_(Al_2O_3` `=` `(10,2)/(102)=0,1` $(mol)$

Theo PT, `n_(NaOH` `=` `2n_(Al_2O_3` `=` `0,2` $(mol)$

Đổi $200$ $ml$ $=$ $0,2$ $l$

`C_MNaOH` `=` `(0,2)/(0,2)=1` $(M)$

$c)$

`m_(ddNaOH` `=` `200.1,15=230` $(g)$

`m_(ddNaAlO_2` `=` `50,2+230-40=240,2` $(g)$

Theo PT, `n_(NaAlO_2` `=` `2n_(Al_2O_3` `=` `0,2` $(mol)$

`m_(NaAlO_2` `=` `0,2.82=16,4` $(g)$

`C%_(NaAlO_2` `=` `(16,4)/(240,2).100%≈6,83%`