$a,PTPƯ:2KClO_3\xrightarrow{t^o} 2KCl+3O_2$
$n_{KClO_3}=\dfrac{6,5}{122,5}≈0,053mol.$
$Theo$ $pt:$ $n_{O_2}=\dfrac{3}{2}n_{KClO_3}=0,0795mol.$
$⇒V_{O_2}=0,0795.22,4=1,7808l.$
$b,PTPƯ:2H_2+O_2\xrightarrow{t^o} 2H_2O$
$n_{H_2}=\dfrac{2,24}{22,4}=0,1mol$
$\text{Lập tỉ lệ:}$ $\dfrac{0,1}{2}<\dfrac{0,0795}{1}$
$⇒O_2$ $dư.$
$Theo$ $pt:$ $n_{H_2O}=n_{H_2}=0,1mol$
$⇒m_{H_2O}=0,1.18=1,8g.$
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