a) \(m_{tăng}=m_{C_2H_4}=0,84\left(g\right)\)

=> \(n_{C_2H_4}=\dfrac{0,84}{28}=0,03\left(mol\right)\)

Gọi số mol CH₄, H₂ trong 3360 ml A là a, b

=> \(a+b=\dfrac{3,36}{22,4}-0,03=0,12\left(mol\right)\) (1)

Gọi số mol CH₄, H₂ trong 0,7 lít hh là ak, bk

=> ak + bk + 0,03k = \(\dfrac{0,7}{22,4}=0,03125\) (2)

Và 16ak + 2bk + 0,84k = 0,4875 (3)

(1)(2)(3) => \(\left\{{}\begin{matrix}a=0,09\left(mol\right)\\b=0,03\left(mol\right)\\k=\dfrac{5}{24}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,03}{0,15}.100\%=20\%\\\%V_{CH_4}=\dfrac{0,09}{0,15}.100\%=60\%\\\%V_{H_2}=\dfrac{0,03}{0,15}.100\%=20\%\end{matrix}\right.\)

b) 

\(n_{C_2H_4}=\dfrac{1,68}{22,4}.20\%=0,015\left(mol\right)\)

\(n_{CH_4}=\dfrac{1,68}{22,4}.60\%=0,045\left(mol\right)\)

\(n_{H_2}=\dfrac{1,68}{22,4}.20\%=0,015\left(mol\right)\)

Bảo toàn C: \(n_{CO_2}=0,075\left(mol\right)\)

Bảo toàn H: \(n_{H_2O}=0,135\left(mol\right)\)

\(n_{Ca\left(OH\right)_2}=0,05.1=0,05\left(mol\right)\)

\(m_{ddCa\left(OH\right)_2}=1000.1,025=1025\left(g\right)\)

PTHH: Ca(OH)₂ + CO₂ --> CaCO₃ + H₂O

             0,05---->0,05----->0,05

            CaCO₃ + CO₂ + H₂O --> Ca(HCO₃)₂

           0,025<--0,025------------>0,025

\(m_{CaCO_3}=\left(0,05-0,025\right).100=2,5\left(g\right)\)

m_{dd sau pư} = 1025 + 0,075.44 + 0,135.18 - 2,5 = 1028,23 (g)

\(C\%_{Ca\left(HCO_3\right)_2}=\dfrac{0,025.162}{1028,23}.100\%=0,3939\%\)

Gọi số mol CO, CH₄ là a, b (mol)

=> \(a+b=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: 2CO + O₂ --t^o--> 2CO₂

              a--->0,5a

            CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

             b--->2b

=> 0,5a + 2b = 0,2

=> a = 0,2 (mol); b = 0,05 (mol)

=> \(\left\{{}\begin{matrix}\%V_{CO}=\dfrac{0,2}{0,25}.100\%=80\%\\\%V_{CH_4}=\dfrac{0,05}{0,25}.100\%=20\%\end{matrix}\right.\)

\(n_{hh}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ 2CO+O_2\rightarrow\left(t^o\right)2CO_2\\ CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ Đặt:n_{CO}=a\left(mol\right);n_{CH_4}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}a+b=0,25\\0,5a+2b=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,05\end{matrix}\right.\\ \Rightarrow\%V_{\dfrac{CO}{hh}}=\%n_{\dfrac{CO}{hh}}=\dfrac{a}{a+b}.100\%=\dfrac{0,2}{0,25}.100=80\%;\%V_{CH_4}=100\%-80\%=20\%\)

a, Ta có \(n_{CH_4}+n_{C_2H_4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\left(1\right)\)

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

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

Theo PT: \(n_{O_2}=2n_{CH_4}+3n_{C_2H_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,05\left(mol\right)\\n_{C_2H_4}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}V_{CH_4}=0,05.22,4=1,12\left(l\right)\\V_{C_2H_4}=0,1.22,4=2,24\left(l\right)\end{matrix}\right.\)

b, \(m_{CH_4}=0,05.16=0,8\left(g\right)\)

c, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

Theo PT: \(n_{Br_2}=n_{C_2H_4}=0,1\left(mol\right)\Rightarrow V_{ddBr_2}=\dfrac{0,1}{1}=0,1\left(l\right)\)

\(m_{C_2H_4}=0,1.28=2,8\left(g\right)\)

tỉ khối là 13 chứ

nX = 0,25

 n↓ = nAnkin = 0,25.20% = 0,05

=>M↓ = 147: C3H3Ag

->Ankin là C3H4 (0,05 mol)
nY = 0,2; nCO2 = 0,3 

=>Số C = nCO2/nY = 1,5

 Ankan là CH4
mY = mX – mC3H4 = 4,5

=>nH2O = \(\dfrac{4,5}{0,3.12}.\dfrac{1}{2}\) = 0,45
 nCH4 = nH2O – nCO2 = 0,15 và nCxH2x = nY – nCH4 = 0,05

nCO2 = 0,15.1 + 0,05x = 0,3

=>x = 3
=> Anken là C3H6

Vì: %m_CH4 = 80%

\(\Rightarrow m_{CH_4}=25.80\%=20\left(g\right)\Rightarrow n_{CH_4}=\dfrac{20}{16}=1,25\left(mol\right)\)

\(\Rightarrow m_{H_2}=5\left(g\right)\Rightarrow n_{H_2}=\dfrac{5}{2}=2,5\left(mol\right)\)

PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)

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

Theo PT: \(\Sigma n_{O_2}=\dfrac{1}{2}n_{H_2}+2n_{CH_4}=3,75\left(mol\right)\)

\(\Rightarrow V_{O_2}=3,75.22,4=84\left(l\right)\)

Mà: %V_O2 = 20%

\(\Rightarrow V_{kk}=\dfrac{84}{20\%}=420\left(l\right)\)

Bạn tham khảo nhé!

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

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

a, Giả sử: \(\left\{{}\begin{matrix}n_{CH_4}=x\left(mol\right)\\n_{C_2H_2}=y\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow x+y=\dfrac{22,4}{22,4}=1\left(mol\right)\left(1\right)\)

Ta có: \(n_{CO_2}=\dfrac{35,84}{22,4}=1,6\left(mol\right)\)

Theo PT: \(\Sigma n_{CO_2}=n_{CH_4}+2n_{C_2H_2}\)

\(\Rightarrow x+2y=1,6\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,4\left(mol\right)\\y=0,6\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,4}{1}.100\%=40\%\\\text{ }\%V_{C_2H_2}=60\%\end{matrix}\right.\)

b, Theo PT: \(\Sigma n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_2}=2,3\left(mol\right)\)

\(\Rightarrow m_{O_2}=2,3.32=73,6\left(g\right)\)

c, PT: \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

Theo PT: \(n_{Na_2CO_3}=n_{CO_2}=1,6\left(mol\right)\)

\(\Rightarrow C_{M_{Na_2CO_3}}=\dfrac{1,6}{0,8}=2M\)

Bạn tham khảo nhé!

a) PTTH: \(2H_2+O_2\rightarrow2H_2O\)

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

b) Ta có: \(\left\{{}\begin{matrix}\overline{M}_{hhkhí}=0,5\cdot28=14\\n_{hhkhí}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\end{matrix}\right.\)

Theo phương pháp đường chéo, ta có: \(\dfrac{n_{H_2}}{n_{C_2H_2}}=\dfrac{12}{12}=1\)

\(\Rightarrow n_{H_2}=n_{C_2H_2}=0,3\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{H_2}=\dfrac{0,3}{0,6}\cdot100\%=50\%\\\%V_{C_2H_2}=50\%\\\%m_{H_2}=\dfrac{0,3\cdot2}{5,6}\cdot100\%\approx10,71\%\\\%m_{C_2H_4}=89,29\%\end{matrix}\right.\)

ta có dhh/H2=6--->Mtrung bình hỗ hợp khí =6*2=12

bạn dùng sơ đồ đường chéo mk làm mẫu phần a kau còn lại tương tự nha

nN2 (28)                        12-2=10

nH2 (2)             12       28-12=16     ta có % thể tich = % về sô mol ----> %Vh2=\(\frac{16}{10+16}\)=61,5 %( xấp xỉ nha) %VN2 = 100-61,5=38,5%