Question

Answer 1

a) \(n_{Br_2}=\dfrac{136.20\%}{160}=0,17\left(mol\right)\)

PTHH: C₃H₆ + Br₂ --> C₃H₆Br₂

         0,17<---0,17

=> \(n_{C_2H_6}=\dfrac{11,04-0,17.42}{30}=0,13\left(mol\right)\)

=> \(\left\{{}\begin{matrix}\%V_{C_2H_6}=\dfrac{0,13}{0,13+0,17}.100\%=43,33\%\\\%V_{C_3H_6}=\dfrac{0,17}{0,13+0,17}.100\%=56,67\%\end{matrix}\right.\)

b) 

Bảo toàn C; \(n_{CaCO_3}=n_C=0,17.3+0,13.2=0,77\left(mol\right)\)

=> \(m_{CaCO_3}=0,77.100=77\left(g\right)\)