1)

Coi \(n_X = 1(mol)\)

Gọi : \(n_{CO_2} = a(mol) ; n_{N_2} = b(mol)\)

Ta có :

\(n_X = a + b = 1(mol)\\ m_X = 44a + 28b = 1.1,225.32(gam)\\ \Rightarrow a = 0,7 ; b = 0,3\)

Vậy :

\(\%V_{CO_2} = \dfrac{0,7}{1}.100\% = 70\%\\ \%V_{N_2} = 100\% - 70\% = 30\%\)

2)

\(n_X = \dfrac{1}{22,4}(mol)\\ \Rightarrow m_X = n.M = \dfrac{1}{22,4}.1,225.32 = 1,75(gam)\)

Gọi số mol O₂, CO₂ là a, b

Có: \(\overline{M}=\dfrac{32a+44b}{a+b}=19,5.2=39\)

=> \(a=\dfrac{5}{7}b\)

=> \(\left\{{}\begin{matrix}\%V_{O_2}=\dfrac{a}{a+b}.100\%=\dfrac{\dfrac{5}{7}b}{\dfrac{5}{7}b+b}.100\%=41,67\%\\\%V_{CO_2}=\dfrac{b}{a+b}.100\%=\dfrac{b}{\dfrac{5}{7}b+b}.100\%=58,33\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{O_2}=\dfrac{32a}{32a+44b}.100\%=34,188\%\\\%m_{CO_2}=\dfrac{44b}{32a+44b}.100\%=65,812\%\end{matrix}\right.\)

\(M_{hh}=19,5.M_{H_2}=19,5.2=39\left(\dfrac{g}{mol}\right)\\ Đặt:a=V_{\dfrac{O_2}{hh}}\\ M_{hh}=39\left(\dfrac{g}{mol}\right)\\ \Leftrightarrow\dfrac{32.a+44.\left(100\%-a\right)}{100\%}=39\\ \Leftrightarrow a=\dfrac{5}{12}\\ \Rightarrow\%V_{\dfrac{O_2}{hh}}=\dfrac{5}{12}.100\%=41,667\%\Rightarrow\%V_{\dfrac{CO_2}{hh}}\approx58,333\%\\ \%m_{\dfrac{O_2}{hh}}=\dfrac{\dfrac{5}{12}.32}{\dfrac{5}{12}.32+\dfrac{7}{12}.44}.100\approx34,188\%\\ \Rightarrow\%m_{\dfrac{CO_2}{hh}}\approx65,812\%\)

Đặt x và y là số mol O 3  và  O 2  có trong 1 mol hỗn hợp khí

Hỗn hợp khí A : (48x + 32y)/(x+y) = 19,2 x 2 = 38,4

→ 3x = 2y → 40% O 3  và 60%  O 2

Đặt x và y là số mol  H 2  và CO có trong 1 mol hỗn hợp khí

Hỗn hợp khí B : (2x + 28y)/(x+y) = 3,6 x 2 = 7,2

→ x = 4y → 80%  H 2  và 20% CO

Gọi \(\left\{{}\begin{matrix}n_{SO_2}=a\left(mol\right)\\n_{SO_3}=b\left(mol\right)\end{matrix}\right.\)

\(d_A\)_/O2=2,25\(\Rightarrow M_A=72đvC\)

\(\Rightarrow\dfrac{64a+80b}{a+b}=72\)

\(\Rightarrow64a+80b=72a+72b\)

\(\Rightarrow a=b\)

\(\%V_{SO_2}=\%V_{SO_3}=\dfrac{a}{a+b}\cdot100\%=50\%\)

\(\%m_{SO_2}=\dfrac{64a}{64a+80b}\cdot100\%=\dfrac{64a}{64a+80b}\cdot100\%=44,44\%\)

\(\%m_{SO_3}=100\%-44,44\%=55,56\%\)

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)

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