1/
$a,PTPƯ:3Fe+2O_2\xrightarrow{t^o} Fe_3O_4$
$b,n_{Fe}=\dfrac{126}{56}=2,25mol.$
$Theo$ $pt:$ $n_{O_2}=\dfrac{2}{3}n_{Fe}=1,5mol.$
$⇒V_{O_2}=1,5.22,4=33,6l.$
$c,PTPƯ:2KClO_3\xrightarrow{t^o} 2KCl+3O_2$
$Theo$ $pt:$ $n_{KClO_3}=\dfrac{2}{3}n_{O_2}=1mol.$
$⇒m_{KClO_3}=1.122,5=122,5g.$$
2/
$a,PTPƯ:2KMnO_4\xrightarrow{t^o} K_2MnO_4+MnO_2+O_2$
$n_{KMnO_4}=\dfrac{31,6}{158}=0,2mol.$
$Theo$ $pt:$ $n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=0,1mol.$
$⇒V_{O_2}=0,1.22,4=2,24l.$
$b,PTPƯ:4P+5O_2\xrightarrow{t^o} 2P_2O_5$
$n_{P}=\dfrac{12,4}{31}=0,4mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,1}{5}<\dfrac{0,4}{4}$
⇒ Lượng oxi không đủ để đốt cháy hết 12,4g P.
$⇒P$ $dư.$
$⇒n_{P}(dư)=0,4-\dfrac{0,1.4}{5}=0,32mol.$
$⇒m_{P}(dư)=0,32.31=9,92g.$
$c,V_{kk}=V_{O_2}.5=2,24.5=11,2l.$
3/
$a,PTPƯ:2KMnO_4\xrightarrow{t^o} K_2MnO_4+MnO_2+O_2$
$n_{KMnO_4}=\dfrac{94,8}{158}=0,6mol.$
$Theo$ $pt:$ $n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=0,3mol.$
$⇒V_{O_2}=0,3.22,4=6,72l.$
$b,PTPƯ:$
$3Fe+2O_2\xrightarrow{t^o} Fe_3O_4$ $(1)$
$2Zn+O_2\xrightarrow{t^o} 2ZnO$ $(2)$
$n_{Fe}=\dfrac{16,8}{56}=0,3mol.$
$Theo$ $pt1:$ $n_{O_2}=\dfrac{2}{3}n_{Fe}=0,2mol.$
$⇒n_{O_2}(pt2)=0,3-0,2=0,1mol.$
$Theo$ $pt2:$ $n_{Zn}=2n_{O_2}=0,2mol.$
$⇒m_{Zn}=0,2.65=13g.$
4/
$a,PTPƯ:2KMnO_4\xrightarrow{t^o} K_2MnO_4+MnO_2+O_2$
$n_{KMnO_4}=\dfrac{63,2}{158}=0,4mol.$
$Theo$ $pt:$ $n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=0,2mol.$
$⇒V_{O_2}=0,2.22,4=4,48l.$
$b,PTPƯ:2Zn+O_2\xrightarrow{t^o} 2ZnO$
$n_{Zn}=\dfrac{32,5}{65}=0,5mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,5}{2}>\dfrac{0,2}{1}$
$⇒Zn$ $dư.$
⇒ Chất rắn sau pư gồm $Zn$ (dư) và $ZnO$
$⇒n_{Zn}(dư)=0,5-\dfrac{0,2.2}{1}=0,1mol.$
$⇒m_{Zn}(dư)=0,1.65=6,5g.$
$Theo$ $pt:$ $n_{ZnO}=2n_{O_2}=0,4mol.$
$⇒m_{ZnO}=0,4.81=32,4g.$
$⇒a=m_{Zn}(dư)+m_{ZnO}=6,5+32,4=38,9g.$
5/
$a,PTPƯ:2KMnO_4\xrightarrow{t^o} K_2MnO_4+MnO_2+O_2$
$n_{KMnO_4}=\dfrac{94,8}{158}=0,6mol.$
$Theo$ $pt:$ $n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=0,3mol.$
$⇒V_{O_2}=0,3.22,4=6,72l.$
$b,PTPƯ:4Al+3O_2\xrightarrow{t^o} 2Al_2O_3$
$Theo$ $pt:$ $n_{Al}=\dfrac{4}{3}n_{O_2}=0,4mol.$
$⇒m_{Al}(pư)=0,4.27=10,8g.$
$⇒a=\dfrac{10,8}{100\%-10\%}=12g.$
$⇒m_{Al}(dư)=12-10,8=1,2g.$
6/
$a,PTPƯ:2Al+6HCl\xrightarrow{} 2AlCl_3+3H_2↑$
$b,n_{Al}=\dfrac{5,4}{27}=0,2mol.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{3}{2}n_{Al}=0,3mol.$
$⇒V_{H_2}=0,3.22,4=6,72l.$
$c,Theo$ $pt:$ $n_{HCl}=3n_{Al}=0,6mol.$
$⇒m_{HCl}=0,6.36,5=21,9g.$
$⇒m_{ddHCl}=\dfrac{21,9}{14,6\%}=150g.$
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