\(\begin{array}{l}
a)\\
m_{Cu}=m_Y=6,4(g)\\
\to m_{Na}+m_{BaO}=24-6,4=17,6(g)\\
2Na+2H_2O\to 2NaOH+H_2\\
BaO+H_2O\to Ba(OH)_2\\
n_{H_2}=\frac{1,12}{22,4}=0,05(mol)\\
Theo\,PT:\,n_{Na}=2n_{H_2}=0,1(mol)\\
\to m_{Na}=0,1.23=2,3(g)\\
\to m_{BaO}=17,6-2,3=15,3(g)\\
\to\begin{cases} \%m_{Cu}=\frac{6,4}{24}.100\%≈26,67\%\\ \%m_{Na}=\frac{2,3}{24}.100\%≈9,58\%\\ \%m_{BaO}=100-9,58-26,67=63,75\%\end{cases}\\
b)\\
n_{BaO}=\frac{15,3}{153}=0,1(mol)\\
Theo\,PT:\,n_{NaOH}=n_{Na}=0,1(mol);n_{Ba(OH)_2}=n_{BaO}=0,1(mol)\\
m_{dd\,spu}=24+200-6,4-0,05.2=217,5(g)\\
\to\begin{cases} C\%_{NaOH}=\frac{0,1.40}{217,5}.100\%≈1,84\%\\ C\%_{Ba(OH)_2}=\frac{0,1.171}{217,5}.100\%≈7,86\% \end{cases}
\end{array}\)

`#Ly`

$a)$

PTHH:

$(1):$ $2Na$ $+$ $H_2O$ `->` $2NaOH$ $+$ $H_2↑$

$(2):$ $BaO$ $+$ $H_2O$ `->` $Ba(OH)_2$

$n_{H_2}$ $=$ `(V)/(22,4)` $=$ `(1,12)/(22,4)` $=$ $0,05$ $(mol)$

`=>` Chất rắn $Y:$ $Cu$

Theo PT $(1)$, `n_(Na` `=` `2n_(H_2` $=$ $0,1$ $(mol)$

`m_(Na` `=` $n.M=0,1.23=2,3$ $(g)$

`%m_(Na` $=$ `(m_{Na})/(m_{hh}).100%` $=$ `(2,3)/(24).100%≈9,58%`

`%_(Cu` `=` `(m_{Cu})/(m_{hh}).100%=(6,4)/(24).100%≈26,67%`

`%_(BaO` `=` `100-9,58-26,67=63,75%`

$b)$

`m_(BaO` `=` `24-2,3-6,4=15,3` $(g)`

`n_(BaO` `=` `m/M` `=` `(15,3)/(153)` `=` `0,1` $(mol)$

Theo PT $(2)$, `n_(Ba(OH)_2` `=` `n_(BaO` `=` `0,1` $(mol)$

Theo PT $(1)$, `n_(NaOH` `=` `Na` `=` `0,1` $(mol)$

`m_(Ba(OH)_2` `=` `n.M=0,1.171=17,1` $(g)$

`m_(NaOH` `=` `n.M=0,1.40=4` $(g)$

`m_(dd(spư)` `=` `24+200-6,4-0,05.2=217,5` $(g)$

`C%_(Ba(OH)_2` `=` `(m_{Ba(OH)_2})/(m_{dd(spư)}).100%` `=` `(17,1)/(217,5).100%≈7,86%`

`C%_(NaOH` `=` `(m_{NaOH})/(m_{dd(spư)}).100%` `=` `(4)/(217,5).100%≈1,84%`