a) PTHH:
\(C_2H_4+3O_2\xrightarrow[]{t^o}2CO_2+2H_2O\) (1)
\(2C_2H_2+5O_2\xrightarrow[]{t^o}4CO_2+2H_2O\) (2)
\(CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3\downarrow+H_2O\) (3)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\) (4)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\) (5)
b) Gọi \(\left\{{}\begin{matrix}n_{C_2H_4}=x\left(mol\right)\\n_{C_2H_2}=y\left(mol\right)\end{matrix}\right.\)
=> \(x+y=\dfrac{8,4}{22,4}=0,375\left(I\right)\)
Theo PT (1), (2): \(n_{O_2}=3n_{C_2H_4}+\dfrac{5}{2}n_{C_2H_2}\)
=> \(3x+2,5y=\dfrac{22,4}{22,4}=1\left(II\right)\)
Từ (I), (II) => x = 0,125; y = 0,25
BTNT C: \(n_{BaCO_3}=n_{CO_2}=2n_{C_2H_2}+2n_{C_2H_4}=0,75\left(g\right)\)
=> \(a=m_{BaCO_3}=0,75.197=147,75\left(g\right)\)
Ta có: \(b=m_{tăng}=m_{C_2H_2}+m_{C_2H_4}=0,125.28+0,25.26=10\left(g\right)\)
