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 --to--> 2CO2 + 2H2O
\(\dfrac{1}{2}a\)----->\(\dfrac{3}{2}a\)
2C2H2 + 5O2 --to--> 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)
