I.
1/
$a,N_2O_5+H_2O\xrightarrow{} 2HNO_3$
$b,K_2O+H_2O\xrightarrow{} 2KOH$
$c,2Al(OH)_3+3H_2SO_4\xrightarrow{} Al_2(SO_4)_3+6H_2O$
$d,Zn+2HCl\xrightarrow{} ZnCl_2+H_2↑$
$e,CuO+H_2\xrightarrow{t^o} Cu+H_2O$
$f,2KMnO_4\xrightarrow{t^o} K_2MnO_4+MnO_2+O_2$
II.
1/
$a,PTPƯ:Zn+H_2SO_4\xrightarrow{} ZnSO_4+H_2↑$
$b,n_{Zn}=\dfrac{6,5}{65}=0,1mol.$
$Theo$ $pt:$ $n_{H_2}=n_{Zn}=0,1mol.$
$⇒V_{H_2}=0,1.22,4=2,24l.$
$c,Theo$ $pt:$ $n_{H_2SO_4}=n_{Zn}=0,1mol.$
$⇒C\%_{H_2SO_4}=\dfrac{0,1.98}{200}.100\%=4,9\%$
$d,PTPƯ:HgO+H_2\xrightarrow{t^o} Hg+H_2O$
$Theo$ $pt:$ $n_{Hg}=n_{H_2}=0,1mol.$
$⇒m_{Hg}=0,1.201=20,1g.$
2/
$a,PTPƯ:2Na+2H_2O\xrightarrow{} 2NaOH+H_2↑$
$n_{H_2}=\dfrac{3,36}{22,4}=0,15mol.$
$Theo$ $pt:$ $n_{Na}=2n_{H_2}=0,3mol.$
$⇒m_{Na}=0,3.23=6,9g.$
$b,Theo$ $pt:$ $n_{NaOH}=2n_{H_2}=0,3mol.$
$⇒m_{NaOH}=0,3.40=12g.$
c, Đổi 500 ml = 0,5 lít.
$⇒CM_{NaOH}=\dfrac{0,3}{0,5}=0,6M.$
3/
$a,PTPƯ:3Fe+2O_2\xrightarrow{t^o} Fe_3O_4$
$b,n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01mol.$
$Theo$ $pt:$ $n_{Fe}=3n_{Fe_3O_4}=0,03mol.$
$⇒m_{Fe}=0,03.56=1,68g.$
$Theo$ $pt:$ $n_{O_2}=2n_{Fe_3O_4}=0,02mol.$
$c,PTPƯ:2KMnO_4\xrightarrow{t^o} K_2MnO_4+MnO_2+O_2$
$Theo$ $pt:$ $n_{KMnO_4}=2n_{O_2}=0,04mol.$
$⇒m_{KMnO_4}=0,04.158=6,32g.$
4/
$a,PTPƯ:2Al+6HCl\xrightarrow{} 2AlCl_3+3H_2↑$
$b,n_{H_2}=\dfrac{3,36}{22,4}=0,15mol.$
$Theo$ $pt:$ $n_{Al}=\dfrac{2}{3}n_{H_2}=0,1mol.$
$⇒a=m_{Al}=0,1.27=2,7g.$
$c,Theo$ $pt:$ $n_{HCl}=2n_{H_2}=0,3mol.$
Đổi 200 ml = 0,2 lít.
$⇒CM_{HCl}=\dfrac{0,3}{0,2}=1,5M.$
5/
$a,PTPƯ:$
$Fe+2HCl\xrightarrow{} FeCl_2+H_2↑$ $(1)$
$CuO+H_2\xrightarrow{t^o} Cu+H_2O$ $(2)$
$n_{Fe}=\dfrac{5,6}{56}=0,1mol.$
$n_{HCl}=\dfrac{7,3}{36,5}=0,2mol.$
$\text{Lập tỉ lệ:}$ $\dfrac{0,1}{1}=\dfrac{0,2}{2}$
⇒ Cả 2 chất đều phản ứng hết.
$Theo$ $pt1:$ $n_{H_2}=n_{Fe}=0,1mol.$
$⇒V_{H_2}=0,1.22,4=2,24l.$
$b,Theo$ $pt2:$ $n_{CuO}=n_{H_2}=0,1mol.$
$⇒m_{CuO}=0,1.80=8g.$
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