Ta có: \(n_{hh}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\)
m bình tăng = mC2H4 = 2,8 (g) \(\Rightarrow n_{C_2H_4}=\dfrac{2,8}{28}=0,1\left(mol\right)\)
⇒ nC3H8 = 0,4 - 0,1 = 0,3 (mol) ⇒ mC3H8 = 0,3.44 = 13,2 (g)
\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,1}{0,4}.100\%=25\%\\\%V_{C_3H_8}=75\%\end{matrix}\right.\)
PT: \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
\(C_3H_8+5O_2\underrightarrow{t^o}3CO_2+4H_2O\)
Theo PT:
\(n_{O_2}=3n_{C_2H_4}+5n_{C_3H_8}=1,8\left(mol\right)\Rightarrow V_{O_2}=1,8.24,79=44,622\left(l\right)\)
\(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{21\%}=212,5\left(l\right)\)
\(n_{CO_2}=2n_{C_2H_4}+3n_{C_3H_8}=1,1\left(mol\right)\Rightarrow V_{CO_2}=1,1.24,79=27,269\left(l\right)\)