Question

Ai giải giúp em câu 14 đi ạ

Answer 1

a) 

P1:

\(n_{Br_2}=\dfrac{80.20\%}{160}=0,1\left(mol\right)\)

PTHH: C₂H₄ + Br₂ --> C₂H₄Br₂

             0,1<--0,1

=> \(n_{C_2H_4\left(P_1\right)}=0,1\left(mol\right)\)

=> \(m_{C_3H_8\left(P_1\right)}=\dfrac{12,2}{2}-0,1.28=3,3\left(g\right)\)

=> \(n_{C_3H_8\left(P_1\right)}=\dfrac{3,3}{44}=0,075\left(mol\right)\)

=> \(V=\left(0,1.2+0,075,2\right).22,4=7,84\left(l\right)\)

\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,1}{0,1+0,075}.100\%=57,143\%\\\%V_{C_3H_8}=\dfrac{0,075}{0,1+0,075}.100\%=42,857\%\end{matrix}\right.\)

b) P₂ \(\left\{{}\begin{matrix}C_2H_4:0,1\left(mol\right)\\C_3H_8:0,075\left(mol\right)\end{matrix}\right.\)

Bảo toàn C: \(n_{CO_2}=0,425\left(mol\right)\) => \(n_{BaCO_3}=0,425\left(mol\right)\)

Bảo toàn H: \(n_{H_2O}=0,5\left(mol\right)\)

Xét \(\Delta m=m_{CO_2}+m_{H_2O}-m_{BaCO_3}=0,425.44+0,5.18-0,425.197=-56,025\left(g\right)\)

=> khối lượng dd sau pư giảm 56,025 gam