Question

Answer 1

Gọi \(\left\{{}\begin{matrix}n_{C_2H_4}=a\left(mol\right)\\n_{C_2H_2}=b\left(mol\right)\end{matrix}\right.\left(đk:a,b>0\right)\)

- Xét phần 1:

\(m_{tăng}=m_{C_2H_4}+m_{C_2H_2}=0,68\left(g\right)\\ \rightarrow\dfrac{1}{2}\left(28a+26b\right)=0,68\\ \Leftrightarrow14a+13b=0,68\left(1\right)\)

- Xét phần 2:

\(n_{O_2}=\dfrac{1,568}{22,4}=0,07\left(mol\right)\)

PTHH:

C₂H₄ + 3O₂ --t^o--> 2CO₂ + 2H₂O

\(\dfrac{1}{2}a\)----->\(\dfrac{3}{2}a\)

2C₂H₂ + 5O₂ --t^o--> 4CO₂ + 2H₂O

\(\dfrac{1}{2}b\)----->\(\dfrac{5}{4}b\)

=> Hệ pt \(\left\{{}\begin{matrix}14a+13b=0,68\\\dfrac{3}{2}a+\dfrac{5}{4}b=0,07\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,03\left(mol\right)\\b=0,02\left(mol\right)\end{matrix}\right.\left(TM\right)\)

=> m = 0,03.28 + 0,02.26 = 1,36 (g)