$a,PTPƯ:2Na+2H_2O\xrightarrow{} 2NaOH+H_2↑$ $(1)$
$n_{Na}=\dfrac{6,9}{23}=0,3mol.$
$Theo$ $pt1:$ $n_{H_2}=\dfrac{1}{2}n_{Na}=0,15mol.$
$⇒V_{H_2}=0,15.22,4=3,36l.$
$b,PTPƯ:CuO+H_2\xrightarrow{t^o} Cu+H_2O$ $(2)$
$\text{Lập tỉ lệ:}$ $\dfrac{0,2}{1}>\dfrac{0,15}{1}$
$⇒CuO$ $dư.$
$⇒n_{CuO}(dư)=-0,2-0,15=0,05mol.$
$⇒m_{CuO}(dư)=0,05.80=4g.$
$c,Theo$ $pt1:$ $n_{NaOH}=n_{Na}=0,3mol.$
$⇒m_{NaOH}=0,3.40=12g.$
$⇒m_{ddNaOH}=m_{Na}+m_{H_2O}-m_{H_2}=6,9+200-(0,15.2)=206,6g.$
$⇒C\%_{NaOH}=\dfrac{m_{NaOH}}{m_{ddNaOH}}.100\%=\dfrac{12}{206,6}.100\%=5,8\%$
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