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/

$a,PTPƯ:2Al+6HCl\xrightarrow{} 2AlCl_3+3H_2↑$

$b,n_{Al}=\dfrac{5,4}{27}=0,2mol.$

$Theo$ $pt:$ $n_{H_2}=\dfrac{3}{2}n_{Al}=0,3mol.$

$⇒V_{H_2}=0,3.22,4=6,72l.$

$c,Theo$ $pt:$ $n_{HCl}=3n_{Al}=0,6mol.$

$⇒m_{HCl}=0,6.36,5=21,9g.$

$⇒m_{ddHCl}=\dfrac{21,9}{14,6\%}=150g.$

chúc bạn học tốt!

Đáp án:

 dưới nha bn

Giải thích các bước giải:

 c6:

a. 2Al+6HCl→2AlCl3+3H2

   0,2→ 0,6                     0,3 

b. V H2(đktc)=0,2 x22,4=4,48 l

c.m ct HCl=0,6 x36,5=21,9 g

   m dd  HCl=21,9 x100:14,6 =150 g

c1:

a.3Fe+2O2→Fe3O4

 2,25 →1,5

b.V O2(đktc)=1,5 x22,4=33,6 l

c.2KClO3→2KCl+3O2

         1               ←1,5

m KCLO3=1 x122,5 =122,5 g

................................chúc bn hk tốt.............................