Ta có: \(d_{NO/H_2}=\dfrac{M_{NO}}{M_{H_2}}=\dfrac{30}{2}=15\)

\(d_{CO_2/O_2}=\dfrac{M_{CO_2}}{M_{O_2}}=\dfrac{44}{32}=1,375\)

\(d_{NO_2/CH_4}=\dfrac{M_{NO_2}}{M_{CH_4}}=\dfrac{46}{16}=2,875\)

Bài 2:

a) Vì khối lượng mol của N₂ và CO đều bằng 28 và lớn hơn khối lượng mol của khí metan CH₄ (28>16)

=> \(d_{\dfrac{hhX}{CH_4}}=\dfrac{28}{16}=1,75\)

Hỗn hợp X nhẹ hơn không khí (28<29)

b)

 \(M_{C_2H_4}=M_{N_2}=M_{CO}=28\left(\dfrac{g}{mol}\right)\\ \rightarrow M_{hhY}=28\left(\dfrac{g}{mol}\right)\\ d_{\dfrac{Y}{H_2}}=\dfrac{28}{2}=14\)

c) \(\%V_{NO}=100\%-\left(30\%+30\%\right)=40\%\\ \rightarrow\%n_{CH_4}=40\%\\ Vì:\%m_{CH_4}=22,377\%\\ Nên:\dfrac{30\%.16}{40\%.30+30\%.16+30\%.\left(x.14+16\right)}=22,377\%\\ \Leftrightarrow x=-0,03\)

Sao lại âm ta, để xíu anh xem lại như nào nhé. 

Bài 1:

\(a.\\ d_{\dfrac{SO_2}{O_2}}=\dfrac{64}{32}=2\\ d_{\dfrac{SO_2}{N_2}}=\dfrac{64}{28}=\dfrac{16}{7}\\ d_{\dfrac{SO_2}{SO_3}}=\dfrac{64}{80}=0,8\\ d_{\dfrac{SO_2}{CO}}=\dfrac{64}{28}=\dfrac{16}{7}\\ d_{\dfrac{SO_2}{N_2O}}=\dfrac{64}{44}=\dfrac{16}{11}\\ d_{\dfrac{SO_2}{NO_2}}=\dfrac{64}{46}=\dfrac{32}{23}\\ b.M_{hhA}=\dfrac{1.64+1.32}{1+1}=48\left(\dfrac{g}{mol}\right)\\ d_{\dfrac{hhA}{O_2}}=\dfrac{48}{32}=1,5\)

88 thì không có trong này

1. Ta có: \(d_{\dfrac{A}{O_2}}=\dfrac{M_A}{32}=0,5\left(lần\right)\)

\(\Leftrightarrow M_A=16\left(g\right)\)

Vậy A là nguyên tố oxi (O)

\(\Rightarrow m_O=\dfrac{4,48}{22,4}.16=3,2\left(g\right)\)

2. 

- \(d_{\dfrac{NO}{N_2}}=\dfrac{30}{28}\approx1,07\left(lần\right)\)

- \(d_{\dfrac{NO_2}{N_2}}=\dfrac{46}{28}\approx1,64\left(lần\right)\)

- \(d_{\dfrac{CO_2}{N_2}}=\dfrac{44}{28}\approx1,57\left(lần\right)\)

a)

$n_{Cl_2} :  n_{O_2} = 1 : 2$

Suy ra : 

$\%V_{Cl_2} = \dfrac{1}{1 + 2}.100\% = 33,33\%$
$\%V_{O_2} = \dfrac{2}{1 + 2}.100\% = 66,67\%$

b)

Coi $n_{Cl_2} = 1 (mol) \Rightarrow n_{O_2} = 2(mol)$
$\%m_{Cl_2} = \dfrac{1.71}{1.71 + 2.32}.100\% = 52,59\%$
$\%m_{O_2} = 100\% -52,59\% = 47,41\%$

c)

$M_A = \dfrac{71.1 + 32.2}{1 + 2} = 45(g/mol)$
$d_{A/B} = \dfrac{45}{28} = 1,607$

a/ $d_{CO/N_2}=\dfrac{28}{28}=1$

b/ $d_{CO_2/O_2}=\dfrac{32}{32}=1$

c/ $d_{N_2/H_2}=\dfrac{28}{2}=14$

e/ $d_{H_2S/H_2}=\dfrac{34}{2}=17$

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)`