Gọi nH2 = a mol ; nC3H6 = b mol
Coi nhh X = 1 mol ⇒ a +b = 1 (1)
Gọi Hiệu suất = a
\(M_X=\frac{2a+42a}{a+b}=\frac{2a+42a}{1}=11.2=22\left(2\right)\)
\(PTHH:C_3H_6+H_2\underrightarrow{^{xt,to}}C_3H_8\)
Ban đầu:___b__________a _____(mol)
Phản ứng:__b n__ bn ____bn____(mol)
Sau phản ứng_b-bn __a-bn ___ bn__(mol)
\(n_{hh\left(spu\right)}=n_{H2\left(dư\right)}+n_{C3H8}+n_{C3H6}=a-bn+bn+b-bn=a+b-bn\left(mol\right)\)
\(M_Y=\frac{m_{H2}+m_{C3H8}+m_{C3H6}}{n_{hh}}=\frac{2a+42bn+42a-42bn}{a+b-bn}=\frac{55}{3}.2=\frac{110}{3}\left(3\right)\)
\(\left(1\right)+\left(2\right)+\left(3\right)\Rightarrow\left\{{}\begin{matrix}a=0,5\\b=0,5\\n=8=80\%\end{matrix}\right.\)
Vậy hiệu suất là 80%
