a)

\(V_{H_2} = V_{O_2} = \dfrac{4,48}{2} = 2,24(lít)\)

b)

\(2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ V_{H_2} = 2,24 < 2V_{O_2} = 4,48(lít)\)

Do đó, O₂ dư.

\(V_{O_2\ pư} = \dfrac{1}{2}V_{H_2} = 1,12(lít)\\ \Rightarrow V_A = V_{O_2\ dư} = 2,24 - 1,12 = 1,12(lít)\)

a, Theo bài ra ta có : \(\left\{{}\begin{matrix}V_{H2}+V_{O2}=4,48\\V_{H2}=V_{O2}\end{matrix}\right.\)

\(\Rightarrow V_{O2}=V_{H2}=2,24\left(l\right)\)

b, Ta có : \(n_{O_2}=n_{H_2}=\dfrac{V}{22,4}=0,1\left(mol\right)\)

\(PTHH:2H_2+O_2\rightarrow2H_2O\)

Thấy sau phản ứng O2 dư .

=> \(V_A=V_{O2du}=22,4\left(n_{o2}-n_{O2pu}\right)=1,12\left(l\right)\)

Vậy ..

Giả sử các khí được đo ở điều kiện sao cho 1 mol khí chiếm thể tích 1 lít

Gọi số mol CH₄, C₂H₆ là a, b (mol)

=> \(a+b=\dfrac{25}{1}=25\left(mol\right)\) (1)

\(n_{O_2}=\dfrac{95}{1}=95\left(mol\right)\)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

             a---->2a---------->a

            2C₂H₆ + 7O₂ --t^o--> 4CO₂ + 6H₂O

              b------>3,5b-------->2b

=> \(\left\{{}\begin{matrix}n_{O_2\left(dư\right)}=95-2a-3,5b\left(mol\right)\\n_{CO_2}=a+2b\left(mol\right)\end{matrix}\right.\)

=> \(95-a-1,5b=\dfrac{60}{1}=60\)

=> a + 1,5b = 35 (2)

(1)(2) => a = 5; b = 20

=> \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{5}{25}.100\%=20\%\\\%V_{C_2H_6}=\dfrac{20}{25}.100\%=80\%\end{matrix}\right.\)

\(\overline{M}_A=\dfrac{5.16+20.30}{5+20}=27,2\left(g/mol\right)\)

\(\overline{M}_B=20,5.2=41\left(g/mol\right)\)

=> \(d_{A/B}=\dfrac{27,2}{41}\approx0,663\)

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

a, \(CH_4+2O_2\underrightarrow{^{t^o}}CO_2+2H_2O\)

\(C_2H_4+3O_2\underrightarrow{^{t^o}}2CO_2+2H_2O\)

b, Gọi: \(\left\{{}\begin{matrix}n_{CH_4}=x\left(mol\right)\\n_{C_2H_4}=y\left(mol\right)\end{matrix}\right.\) \(\Rightarrow x+y=\dfrac{4,48}{22,4}=0,2\left(mol\right)\left(1\right)\)

Theo PT: \(n_{O_2}=2n_{CH_4}+3n_{C_2H_4}=2x+3y=\dfrac{15,68}{22,4}=0,7\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=-0,1\\y=0,3\end{matrix}\right.\)

Đến đây thì ra số mol âm, bạn xem lại đề nhé.

a, \(CH_4+2O_2\underrightarrow{t^o}CO_2+H_2O\)

\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

Ta có: \(n_{CH_4}+n_{C_2H_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\left(1\right)\)

Theo PT: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_2}=\dfrac{56}{22,4}=2,5\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,5\left(mol\right)\\n_{C_2H_2}=1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,5.22,4}{33,6}.100\%\approx33,33\%\\\%V_{C_2H_2}\approx66,67\%\end{matrix}\right.\)

b, Theo PT: \(n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_2}=3,5\left(mol\right)\Rightarrow m_{O_2}=3,5.32=112\left(g\right)\)