$a,PTPƯ:2Na+2H_2O\xrightarrow{} 2NaOH+H_2↑$
$n_{Na}=\dfrac{6,9}{23}=0,3mol.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{Na}=0,15mol.$
$⇒V_{H_2}=0,15.22,4=3,36l.$
$b,PTPƯ:CuO+H_2\xrightarrow{t^o} Cu+H_2O$
$n_{CuO}=\dfrac{3,2}{80}=0,04mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,04}{1}<\dfrac{0,15}{1}$
$⇒H_2$ $dư.$
$⇒n_{H_2}(dư)=0,15-0,04=0,11mol.$
$⇒m_{H_2}(dư)=0,11.2=0,22g.$
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