$\text{#Ephemeral}$

$\text{a,V$_{SO2}$=0,5.22,4=11,2(llits)}$

$\text{b,n$_{CO2}$=2,2:44=0,05(mol)}$

$\text{V$_{CO2}$=0,05.22,4=1,12(Lít)}$

$\text{c,n$_{O2}$=1,6:32=0,05(mol)}$

$\text{n$_{H2}$=2:2=1(mol)}$

$\text{n$_{hh}$=0,05+1=1,05(mol)}$

$\text{Câu 7 :}$

$\text{%O$_{trong_Fe2O3}$=$\frac{3.16}{3.16+2.56}$.100% =30%}$

$\text{PTK: 2.56+16.3=160(đvC)}$

$\text{%O$_{trong_Fe3O4}$=$\frac{4.16}{3.56+4.16}$ .100%≈27,59%}$

$\text{PTK:3.56+16.4=232(đvC)}$

$\text{%O$_{K2O}$=$\frac{16}{16+2.39}$.100% ≈17,02%}$

$\text{PTK:2.39+16=94(đvC)}$

Câu 8:

$\text{Gọi CTHH của h/c là C$_{x}$H$_{y}$O$_{z}$}$

$\text{Theo bài ra ta có :}$

$\text{x:y:z=n$_{C}$:n$_{H}$:n$_{O}$}$

$\text{=$\frac{m_C}{12}$: $\frac{m_H}{1}$ :$\frac{m_O}{16}$ }$

=$\frac{40%}{12}$: $\frac{6,6%}{1}$ :$\frac{(100-40-6,67)%}{16}$

=3,33:6,67:3,33

=1:2:1

Vậy công thức đơn giản nhất của X là CH₂O.