a) CH₄ + 2O₂  → CO₂  + 2H₂O

C₂H₄  + 3O₂ → 2CO₂ + 2H₂O

b) nCO₂ = 4,48/22,4 =0,2 lít

Gọi số mol của CH₄ và C₂H₄ trong hỗn hợp ban đầu lần lượt là x và y ta có:

\(\left\{{}\begin{matrix}16x+28y=3\\x+2y=0,2\end{matrix}\right.\) => x = 0,1 và y = 0,05 (mol)

=> nO₂ = 2nCH₄ + 3nC₂H₄ = 0,35 mol <=> V_O2 = 0,35.22,4 = 7,84 lít

c) mCH₄ = 0,1.16 = 1,6 gam => %mCH₄ =\(\dfrac{1,6}{3}.100\)=53,34% , %mC₂H₄ = 100- 53,34 = 46,67%.

Phần trăm về thể tích bằng phần trăm về số mol.

%V_CH4 = \(\dfrac{0,1}{0,15}.100\)= 66,67%   => %V_C2H4 = 100- 66,67 = 33,33%

nX = 0,672/22,4 = 0,03 (mol)

Gọi nN2 = a (mol); nO2 = b (mol)

a + b = 0,03

28a + 32b = 0,88 

=> a = 0,02 (mol); b = 0,01 (mol)

%VN2 = 0,02/0,03 = 66,66%

%VO2 = 100% - 66,66% = 33,34%

M(X) = 0,88/0,03 = 88/3 (g/mol)

nX = 2,2 : 88/3 = 0,075 (mol)

VH2 = VX = 0,075 . 22,4 = 1,68 (l)

Gọi \(\left\{{}\begin{matrix}n_{C_2H_2}=x\\n_{CH_4}=y\end{matrix}\right.\)

\(n_{hh}=\dfrac{5,6}{22,4}=0,25mol\)

Ta có:

\(\left\{{}\begin{matrix}x+y=0,25\\26x+16y=5,5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{0,15}{0,25}.100=60\%\\\%V_{CH_4}=100\%-60\%=40\%\end{matrix}\right.\)

ko bit

nà ní

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

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

\(\left\{{}\begin{matrix}SO_2:64\\O_2:32\end{matrix}\right.\)       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. \(m_{SO_2}=0,3.64=19,2g\)

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

số mol hh X = 0.3(mol) có khối lượng = 8.8 (g)

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

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

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

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)