Question

Answer 1

TN1: Gọi (n_CH4; n_C2H4; n_C2H2) = (a; b; c)

=> \(a+b+c=\dfrac{11,2}{22,4}=0,5\) (1)

m_tăng = 28b + 26c = 8,2 (2)

TN1: Gọi (n_CH4; n_C2H4; n_C2H2) = (ak; bk; ck)

=> \(ak+bk+ck=\dfrac{5,6}{22,4}=0,25\) (3)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

             ak----------------->ak

             C₂H₄ + 3O₂ --t^o--> 2CO₂ + 2H₂O

              bk----------------->2bk

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

                 ck------------------>2ck

=> \(ak+2bk+2ck=\dfrac{8,96}{22,4}=0,4\) (4)

(1)(2)(3)(4) => \(\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,2\left(mol\right)\\c=0,1\left(mol\right)\\k=0,5\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,2.16}{0,2.16+0,2.28+0,1.26}.100\%=28,07\%\\\%m_{C_2H_4}=\dfrac{0,2.28}{0,2.16+0,2.28+0,1.26}.100\%=49,123\%\\\%m_{C_2H_2}=\dfrac{0,1.26}{0,2.16+0,2.28+0,1.26}.100\%=22,807\%\end{matrix}\right.\)