Đáp án:
\(\%V_{C_xH_y}=60\%\\ \%V_{CO}=40\%\)
CTHH của \(\text C_{\text x}\text H_{\text y}\) là \(\text C_3\text H_4\)
Giải thích các bước giải:
\(\text n_{\text{kk}}=\dfrac{72,8}{22,4}=3,25\ \text{mol}\\\Rightarrow n_{O_2}=3,25\cdot 20\%=0,65\ \text{mol}\\\Rightarrow n_{N_2}=3,25-0,65=2,6\ \text{mol}\\ n_{Y}=\dfrac{5,6}{22,4}=0,25\ \text{mol}\\ n_{CO_2}=\dfrac{24,2}{44}=0,55\ \text{mol}\\ n_{H_2O}=\dfrac{5,4}{18}=0,3\ \text{mol}\)
Bảo toàn mol O: \(n_{CO}+0,65\cdot 2=0,55\cdot 2+0,3\\\Rightarrow n_{CO}=0,1\ \text{mol}\\\Rightarrow n_{C_xH_y}=0,25-0,1=0,15\ \text{mol}\)
\(\to \%V_{C_xH_y}=\dfrac{0,15}{0,25}\cdot 100\%=60\%\to \%V_{CO}=100\%-60\%=40\%\)
PTHH: \(CO+\dfrac 12O_2\xrightarrow{t^{\circ}} CO_2\\ \ 0,1\to\ \ \ \ \ \ \ \ \ \ \ \ \ \ 0,1\\ C_xH_y+\left(x+\dfrac y4\right)O_2\xrightarrow{t^{\circ}} xCO_2+\dfrac y2H_2O\\ 0,15\to\qquad\qquad\qquad\ \ \ \ 0,15x\ \ \ 0,075y\)
Suy ra:
\(0,1+0,15x=0,55\) `<=> x=3`
\(0,075y=0,3\) `<=> y=4`
Vậy CTHH cần tìm là \(C_3H_4\)
