Đáp án:

Giải thích các bước giải:

 9C

10D

11C

12A

14D

15D

16B

17B

18B

19B

20C

21C

22B

23C

24A

Tự luận

BT1

a) $S+O2--to-->SO2$

b) $3Fe+2O2--to-->Fe3O4$

c) $2KClO3--to-->2KCl+3O2$

d) $CH4+2O2--->CO2+2H2O$

BT2

a) $4P+5O2--to-->2P2O5$

b)

$nP=7,75/31=0,25(mol)$

Theo PTHH

$nP2O5=1/2nP=0,125(mol)$

=>$mP2O5=0,125.142=17,75(g)$

c)

Theo PTHH

$nO2=5/4 nP=0,3125(mol)$

=>$VO2=0,3125.22,4=7(l)$

Do $Vkk=5VO2$

=>$Vkk=7.5=35(l)$

Đáp án:

Giải thích các bước giải:

$\text{I.Trắc nghiệm.}$

$9.C$

$PTPƯ:2KMnO_4\buildrel{{t^o}}\over\longrightarrow$ $K_2MnO_4+MnO_2+O_2$

$n_{KMnO_4}=\frac{31,6}{158}=0,2mol.$

$Theo$ $pt:$ $n_{O_2}=\frac{1}{2}n_{KMnO_4}=0,1mol.$

$⇒V_{O_2}=0,1.22,4=2,24l.$

$10.D$

$PTPƯ:2KClO_3\buildrel{{t^o}}\over\longrightarrow$ $2KCl+3O_2$

$Theo$ $pt:$ $n_{KClO_3}=\frac{2}{3}n_{O_2}=0,2mol.$

$⇒m_{KClO_3}=0,2.122,5=24,5g.$

$11.C$

$12.A$

$13.D$

$14.B$

$15.D$

$16.B$

$17.B$

$18.B$

$19.B$

$20.C$

$21.C$

$22.B$

$23.C$

$24.A$

$\text{II.Tự luận.}$

$1/$

$a,S+O_2\buildrel{{t^o}}\over\longrightarrow$ $SO_2$

$b,3Fe+2O_2\buildrel{{t^o}}\over\longrightarrow$ $Fe_3O_4$

$c,2KClO_3\buildrel{{đp}}\over\longrightarrow$ $2KCl+3O_2$

$d,CH_4+2O_2\buildrel{{t^o}}\over\longrightarrow$ $CO_2+H_2O$

$2/$

$a,PTPƯ:4P+5O_2\buildrel{{t^o}}\over\longrightarrow$ $2P_2O_5$

$b,n_{P}=\frac{7,75}{31}=0,25mol.$

$Theo$ $pt:$ $n_{P_2O_5}=\frac{1}{2}n_{P}=0,125mol.$

$⇒m_{P_2O_5}=0,125.142=17,75g.$

$c,Theo$ $pt:$ $n_{O_2}=\frac{5}{4}n_{P}=0,3125mol.$

$⇒V_{O_2}=0,3125.22,4=7l.$

$\text{Vì VO2=20% Vkk nên:}$

$⇒V_{kk}=7.5=35l.$

chúc bạn học tốt!