a) \(n_{CH_4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

          0,05-->0,1------->0,05

             2C₂H₂ + 5O₂ --t^o--> 4CO₂ + 2H₂O

            0,125<--0,3125<----0,25

=> \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,05}{0,05+0,125}.100\%=28,57\%\\\%V_{C_2H_2}=\dfrac{0,125}{0,05+0,125}.100\%=71,43\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,05.16}{0,05.16+0,125.26}.100\%=19,753\%\\\%m_{C_2H_2}=\dfrac{0,125.26}{0,05.16+0,125.26}.100\%=80,247\%\end{matrix}\right.\)

b) \(n_{O_2}=0,1+0,3125=0,4125\left(mol\right)\)

=> \(V_{O_2}=0,4125.22,4=9,24\left(l\right)\)

=> V_kk = 9,24.5 = 46,2 (l)

\(a.C_2H_5OH+3O_2-^{t^o}\rightarrow2CO_2+3H_2O\\ n_{C_2H_5OH}=0,3\left(mol\right)\\ n_{CO_2}=2n_{C_2H_5OH}=0,6\left(mol\right)\\ \Rightarrow V_{CO_2}=0,6.22,4=13,44\left(l\right)\\ b.n_{O_2}=3n_{C_2H_5OH}=0,6\left(mol\right)\\ MàV_{O_2}=\dfrac{1}{5}V_{kk}\\ \Rightarrow V_{kk}=V_{O_2}.5=0,6.22,4.5=67,2\left(l\right)\\ c.n_{NaOH}=0,9\left(mol\right)\\ Tacó:\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,9}{0,6}=1,5\\ \Rightarrow Tạora2muốiNaHCO_3vàNa_2CO_3\\ Đặt:n_{NaHCO_3}=x\left(mol\right);n_{Na_2CO_3}=y\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}x+y=0,6\left(BTnguyento\left(C\right)\right)\\x+2y=0,9\left(BTnguyento\left(Na\right)\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,3\\y=0,3\end{matrix}\right.\\ \Rightarrow m_{muối}=0,3.84+0,3.106=57\left(g\right)\)

a)C2H6O+3O2to→2CO2+3H2O

0,3-------------0,9------0,6

nC2H6O=\(\dfrac{13,8}{46}\)=0,3mol

nCO2=2nC2H6O=0,6mol

⇒VCO2=0,6×22,4=13,44l

⇒Vkk=0,9×22,4×5=100,8l

\(n_{C_2H_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

\(n_{O_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)

      \(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{t^0}2CO_2+H_2O\)

\(Bđ:0.3.......0.5\)

\(Pư:0.2........0.5.........0.4.........0.2\)

\(Kt:0.1..........0..........0.4...........0.2\)

\(V_{CO_2}=0.4\cdot22.4=8.96\left(l\right)\)

\(V_{C_2H_2\left(dư\right)}=0.1\cdot22.4=2.24\left(l\right)\)

\(n_{C_2H_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)

\(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{^{^{t^0}}}2CO_2+H_2O\)

\(0.05.......0.125........0.1\)

\(V_{CO_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(V_{kk}=5\cdot V_{O_2}=5\cdot0.125\cdot22.4=14\left(l\right)\)

\(n_{CO_2}=\dfrac{4.4}{44}=0.1\left(mol\right)\)

\(CH_4+2O_2\underrightarrow{^{^{t^0}}}CO_2+2H_2O\)

\(0.1.......0.2........0.1..........0.2\)

\(m_{CH_4}=0.1\cdot16=1.6\left(g\right)\)

\(V_{H_2O}=0.2\cdot22.4=4.48\left(l\right)\)

\(V_{kk}=5V_{O_2}=5\cdot0.2\cdot22.4=22.4\left(l\right)\)

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

\(n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: C₂H₄ + 3O₂ --t^o--> 2CO₂ + 2H₂O

             0,25<--0,75<------0,5

=> \(\left\{{}\begin{matrix}m_{C_2H_4}=0,25.28=7\left(g\right)\\V_{O_2}=0,75.22,4=16,8\left(l\right)\end{matrix}\right.\)

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

b) Vì trong chất khí, tỉ lệ về số mol cũng chính là tỉ lệ về thể tích

\(\Rightarrow V_{CH_4}=V_{CO_2}=3,36\left(l\right)\)