a) \(\left\{{}\begin{matrix}n_{Cl_2}+n_{O_2}=\dfrac{6,72}{22,4}=0,3\\\overline{M}=\dfrac{71.n_{Cl_2}+32.n_{O_2}}{n_{Cl_2}+n_{O_2}}=2.29=58\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}n_{Cl_2}=0,2\left(mol\right)\\n_{O_2}=0,1\left(mol\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%V_{Cl_2}=\dfrac{0,2}{0,3}.100\%=66,67\%\\\%V_{O_2}=\dfrac{0,1}{0,3}.100\%=33,33\%\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}m_{Cl_2}=0,2.71=14,2\left(g\right)\\m_{O_2}=0,1.32=3,2\left(g\right)\end{matrix}\right.\)

Giả sử: \(\left\{{}\begin{matrix}n_{N_2O}=x\left(mol\right)\\n_{O_2}=y\left(mol\right)\end{matrix}\right.\) (trong 12 gam hỗn hợp B)

⇒ 44x + 32y = 12 (1)

Theo đầu bài: d_B/CH4 = 2,5

⇒ M_B = 2,5.16 = 40 (g/mol)

\(\Rightarrow\dfrac{44x+32y}{x+y}=40\Leftrightarrow4x-8y=0\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}V_{N_2O}=0,2.22,4=4,48\left(l\right)\\V_{O_2}=0,1.22,4=2,24\left(l\right)\end{matrix}\right.\)

Bạn tham khảo nhé!

a. Gọi x, y lần lượt là số mol của CH₄ và CO₂

Ta có: \(n_A=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Theo đề, ta có:

- x + y = 0,4 (1)

- 16x + 44y = 9,2 (2)

Từ (1) và (2), ta có HPT: 

\(\left\{{}\begin{matrix}x+y=0,4\\16x+44y=9,2\end{matrix}\right.\)

Giải ra, ta được: 

 x = 0,3, y = 0,1

=> \(m_{CH_4}=0,3.16=4,8\left(g\right);m_{CO_2}=0,1.44=4,4\left(g\right)\)

b. Ta có: \(\overline{M_A}=\dfrac{4,8+4,4}{0,3+0,1}=23\left(g\right)\)

=> \(d_{\dfrac{A}{O_2}}=\dfrac{\overline{M_A}}{M_{O_2}}=\dfrac{23}{32}=0,71875\left(lần\right)\)

Giả sử có 1 mol khí Cl₂, 2 mol khí O₂

a) \(\left\{{}\begin{matrix}\%V_{Cl_2}=\dfrac{1}{1+2}.100\%=33,33\%\\\%V_{O_2}=\dfrac{2}{1+2}.100\%=66,67\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{Cl_2}=\dfrac{1.71}{1.71+2.32}.100\%=52,59\%\\\%m_{O_2}=\dfrac{2.32}{1.71+2.32}.100\%=47,41\%\end{matrix}\right.\)

b) \(\overline{M}=\dfrac{1.71+2.32}{1+2}=45\left(g/mol\right)\)

=> \(d_{A/H_2}=\dfrac{45}{2}=22,5\)

c) \(n_A=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

=> m_A = 0,3.45 = 13,5 (g)

\(n_{hh}=\dfrac{13,44}{22,4}=0,6mol\)

\(\overline{M_x}=24.2=48\)

\(\left\{{}\begin{matrix}SO_2:64\\O_2:32\end{matrix}\right.\)       48    = \(\dfrac{16}{16}=1\)

\(\Rightarrow n_{SO_2=}n_{O_2}=0,3mol\)

1. \(m_{hh}=0,3.64+0,3.32=28,8g\)

2. \(\%V_{SO_2}=\dfrac{0,3.22,4}{13,44}.100\%=50\%\)

\(\Rightarrow\%V_{O_2}=50\%\)

3. \(m_{SO_2}=0,3.64=19,2g\)

\(m_{O_2}=0,3.32=9,6g\)

\(Coi: n_{Cl_2} = 1(mol) \to n_{O_2} = 2(mol)\\ \%V_{Cl_2} = \dfrac{1}{1+2}.100\% = 33,33\%\\ \%V_{O_2} = 100\% -33,33\% = 66,67\%\\ M_A = \dfrac{1.71+2.32}{1+2}=45(g/mol)\\ d_{A/H_2} = \dfrac{45}{2} = 22,5\)

\(\text{Trong 6,72 lít khí A : }m_A = 45.\dfrac{6,72}{22,4}=13,5(gam)\)

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

Áp dụng quy tắc đường chéo:

\(a.\\ \Rightarrow\dfrac{V_{Cl_2}}{V_{O_2}}=\dfrac{15,6}{23,4}=\dfrac{2}{3}\\ \Rightarrow\left\{{}\begin{matrix}\%V_{Cl_2}=40\%\\\%V_{O_2}=60\%\end{matrix}\right.\)

\(b.\) 

Ta có: \(\dfrac{n_{Cl_2}}{n_{O_2}}=\dfrac{2}{3}\Leftrightarrow\dfrac{m_{Cl_2}}{m_{O_2}}=\dfrac{71.2}{32.3}=\dfrac{71}{48}\Leftrightarrow48m_{Cl_2}-71m_{O_2}=0\)

Mặt khác: \(m_{Cl_2}+m_{O_2}=5,95\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Cl_2}=3,55\left(g\right)\\m_{O_2}=2,4\left(g\right)\end{matrix}\right.\)

Gọi số mol O₂, CO₂ là a, b

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

=> \(a=\dfrac{5}{7}b\)

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

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

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