Gọi a, b lần lượt là thể tích của SO₂ và O₂ (a,b>0)

Vì d_(SO2,O2)/H2=24 => \(\overline{M}\)= \(\frac{64a+32b}{a+b}=48\) => a=b (1)
Mà V hh= 20l => a+b=20 (2)

Từ (1) và (2)=> a=b=10 (l)

Gọi x là V_O2 cần thêm
Vì d_{hh mới/H2}=22,4 => \(\overline{M}\)=\(\frac{640+32x+320}{20+x}\)=44,8
=>x=5l
Chúc bạn học tốt

\(M_X=2,5.16=40\)(g/mol)

\(\rightarrow\dfrac{V_{CO_2}}{V_{O_2}}=\dfrac{n_{CO_2}}{n_{O_2}}=\dfrac{40-32}{44-40}=2\)

Mà \(V_{CO_2}+V_{O_2}=30\left(L\right)\)

\(\rightarrow V_{CO_2}=20\left(L\right);V_{O_2}=10\left(L\right)\)

\(\rightarrow M_Y=\dfrac{20.44+10.32+32V}{V+20+10}=2,25.16=36\)

\(\rightarrow V=30\left(L\right)\)

a) \(M_X=19.2=38\left(g/mol\right)\)

`=>` \(d_{X/kk}=\dfrac{38}{29}=1,310345\)

b) \(m_X=0,4.38=15,2\left(g\right)\)

Gọi \(\left\{{}\begin{matrix}n_{O_2}=x\left(mol\right)\\n_{CO_2}=y\left(mol\right)\end{matrix}\right.\)

`=>` \(\left\{{}\begin{matrix}32x+44y=15,2\\x+y=0,4\end{matrix}\right.\Leftrightarrow x=y=0,2\)

\(m_Y=0,1.28+15,2=18\left(g\right)\)

`=>` \(\left\{{}\begin{matrix}\%m_{N_2}=\dfrac{0,1.28}{18}.100\%=15,56\%\\\%m_{O_2}=\dfrac{0,2.32}{18}.100\%=35,56\%\\\%m_{CO_2}=100\%-15,56\%-35,56\%=48,88\%\end{matrix}\right.\)

b) \(M_{hh}=4.10=40\left(g/mol\right)\)

Gọi \(n_{NO_2}=a\left(mol\right)\)

`=>` \(\left\{{}\begin{matrix}m_{hh}=18+46a\left(g\right)\\n_{hh}=0,5+0,1+a=0,6+a\left(mol\right)\end{matrix}\right.\)

`=>` \(M_{hh}=\dfrac{m_{hh}}{n_{hh}}=\dfrac{18+46a}{0,6+a}=40\)

`=> a = 1`

`=> V_{NO_2(đktc)} = 1.22,4 = 22,4 (l)`

Đáp án B