CHÚC BẠN HỌC TỐT

$\text{Đáp án+Giải thích các bước giải:}$

$\text{Câu 25:}$

$\text{PTHH: S+$O_{2}$→$SO_{2}$}$

$\text{Ta có: nS=$\dfrac{3,2}{32}$= 0,1 mol}$

$\text{Theo PTHH ta có: $nO_{2}$=nS= 0,1 mol}$

$\text{⇒VO2=0,1×22,4= 2,24 lít}$

$\text{⇒VKK=2,24÷$\dfrac{1}{5}$= 11,2 lít}$

$\text{⇒ Chọn A}$

$\text{Câu 26:}$

$\text{PTHH: C+$O_{2}$→$CO_{2}$}$

$\text{Ta có: nC=$\dfrac{2,4}{12}$= 0,2 mol}$

$\text{Theo PTHH ta có: nO2=nC= 0,2 mol}$

$\text{⇒ $VO_{2}$=0,2×22,4= 4,48 lít}$

$\text{⇒ VKK=4,48÷$\dfrac{1}{5}$= 22,4 lít}$

$\text{⇒ Chọn D}$

$\text{Câu 27:}$

$\text{$MCO_{2}$=12+16×2= 44 g/mol}$

$\text{%mO=$\dfrac{16×2}{44}$×100= 72,72%}$

$\text{$MSO_{3}$=32+16×3= 80 g/mol}$

$\text{%mO=$\dfrac{16×3}{80}$×100= 60%}$

$\text{$MP_{2}O_{5}$=31×2+16×5= 142 g/mol}$

$\text{%mO=$\dfrac{16×5}{142}$×100= 56,34%}$

$\text{$MFe_{3}O_{4}$=56×3+16×4= 232 g/mol}$

$\text{%mO=$\dfrac{16×4}{232}$×100= 27,59%}$

$\text{Ta có:}$

$\text{27,59%<56,34%<60%<72,72%}$

$\text{⇔$Fe_{3}O_{4}$<$P{2}O{5}<$SO_{3}<$CO_{2}$}$

$\text{Vậy Oxit chứa hàm lượng Oxi cao nhất là CO2}$

$\text{⇒ Chọn C}$

$\text{Câu 28:}$

$\text{$MNa_{2}O$=23×2+16×= 62 g/mol}$

$\text{%mO=$\dfrac{16}{62}$×100= 25,81%}$

$\text{MCaO=40+16= 56 g/mol}$

$\text{%mO=$\dfrac{16}{56}$×100= 28,57%}$

$\text{$MK_{2}O$=39×2+16= 94 g/mol}$

$\text{%mO=$\dfrac{16}{94}$×100= 56,34%}$

$\text{MFeO=56+16= 72 g/mol

$\text{%mO=$\dfrac{16}{72}$×100= 22,22%}$

$\text{Ta có:}$

$\text{22,22%<25,81%<28,57%<56,34%}$

$\text{⇔FeO<$Na_{2}O$<CaO<$K_{2}O$}$

$\text{Vậy Oxit chứa hàm lượng oxi thấp nhất là FeO}$

$\text{⇒ Chọn A}$

$\text{Câu 29:}$

$\text{PTHH: C+O2→CO2}$

$\text{Ta có: $\dfrac{1,2}{12}$= 0,1 mol}$

$\text{Theo PTHH ta có: nO2=nC= 0,1 mol}$

$\text{PTHH: $2KMnO_{4}$+$O_{2}$→$K_{2}MnO_{4}$+$O_{2}$+$MnO_{2}$}$

$\text{Theo PTHH ta có: $nKMnO_{4}$=2×0,1= 0,2 mol}$

$\text{⇒ $mKMnO_{4}$=0,2×158= 31,6 g}$

$\text{⇒ Chọn A}$

$\text{Câu 30:}$

$\text{PTHH: $2KMnO_{4}$+$O_{2}$→$K_{2}MnO_{4}$+$O_{2}$+$MnO_{2}$}$

$\text{Theo PTHH ta có: $\dfrac{1}{2}$×1= 0,5 mol}$

$\text{⇒ VO2=0,5×22,4= 11,2 lít}$

$\text{PTHH: $2KClO_{3}$→2KCl+$3O_{2}$}$

$\text{Theo PTHH ta có: nO2=$\dfrac{3}{2}$×1= 1,5 mol}$

$\text{⇒ VO2=1,5×22,4= 33,6 lít}$

$\text{Ta có: V1<V2}$

$\text{⇒ Chọn C}$

$\text{Câu 31:}$

$\text{Cây xanh có khả năng hút khí CO2 và nhả khí O}$

$\text{⇒ Chọn D}$

$\text{Câu 32:}$

$\text{PTHH: C+$O_{2}$→$CO_{2}$}$

$\text{Theo PTHH ta có: $nO_{2}$=nC= 0,1 mol}$

$\text{⇒ VO2=0,1×22,4= 2,24 lít}$

$\text{PTHH: S+$O_{2}$→$SO_{2}$}$

$\text{Theo PTHH ta có: $O_{2}$=nS= 0,1 mol}$

$\text{⇒ VO2=0,1×22,4= 2,24 lít}$

$\text{Ta có: V1=V2}$

$\text{⇒ Chọn A}$

$\text{Câu 33:}$

$\text{PTHH: C+$O_{2}$→$CO_{2}$}$

$\text{Ta có: nC=$\dfrac{1}{12}$= 0,08 mol}$

$\text{Theo PTHH ta có: $nO_{2}$=nC= 0,08 mol}$

$\text{⇒ VO2=0,08×22,4= 1,792 lít}$

$\text{PTHH: S+$O_{2}$→$SO_{2}$}$

$\text{Ta có: nS=$\dfrac{1}{12}$= 0,08 mol}$

$\text{Theo PTHH ta có: $O_{2}$=nS= 0,08 mol}$

$\text{⇒ VO2=0,08×22,4= 1,792 lít}$

$\text{Ta có: V1=V2}$