1/
$a,PTPƯ:2KClO_3\xrightarrow{t^o} 2KCl+3O_2$
$n_{KClO_3}=\dfrac{24,5}{122,5}=0,2mol.$
$Theo$ $pt:$ $n_{O_2}=\dfrac{3}{2}n_{KClO_3}=0,3mol.$
$⇒V_{O_2}=0,3.22,4=6,72l.$
$b,PTPƯ:2H_2+O_2\xrightarrow{t^o} 2H_2O$
$n_{H_2}=\dfrac{8,96}{22,4}=0,4mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,4}{2}<\dfrac{0,3}{1}$
$⇒O_2$ $dư.$
$Theo$ $pt:$ $n_{H_2O}=n_{H_2}=0,4mol.$
$⇒m_{H_2O}=0,4.18=7,2g.$
2/
$a,PTPƯ:Zn+H_2SO_4\xrightarrow{} ZnSO_4+H_2↑$
$n_{Zn}=\dfrac{13}{65}=0,2mol.$
$m_{H_2SO_4}=200.15\%=30g.$
$⇒n_{H_2SO_4}=\dfrac{30}{98}=0,3mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,2}{1}<\dfrac{0,3}{1}$
$Theo$ $pt:$ $n_{H_2}=n_{Zn}=0,2mol.$
$⇒V_{H_2}=0,2.22,4=4,48l.$
$b,n_{H_2SO_4}(dư)=0,3-\dfrac{0,2.1}{1}=0,1mol.$
$⇒C\%_{H_2SO_4}=\dfrac{0,1.98}{13+200-0,2.2}.100\%=4,6\%$
$Theo$ $pt:$ $n_{ZnSO_4}=n_{Zn}=0,2mol.$
$⇒C\%_{ZnSO_4}=\dfrac{0,2.161}{13+200-0,2.2}.100\%=15,09\%$
$c,PTPƯ:CuO+H_2\xrightarrow{t^o} Cu+H_2O$
$n_{CuO}=\dfrac{12}{80}=0,15mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,15}{1}<\dfrac{0,2}{1}$
$⇒H_2$ $dư.$
$Theo$ $pt:$ $n_{Cu}=n_{CuO}=0,15mol.$
$⇒m_{Cu}=0,15.64=9,6g.$
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