Đáp án:
\(CTPT:{C_3}{H_8}\)
Giải thích các bước giải:
\(\begin{array}{l}
2{C_x}{H_y} + (4x + y){O_2} \to 2xC{O_2} + y{H_2}O\\
{n_{C{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
{n_{{H_2}O}} = \dfrac{m}{M} = \dfrac{{7,2}}{{18}} = 0,4mol\\
{n_C} = {n_{C{O_2}}} = 0,3mol\\
{n_H} = 2{n_{{H_2}O}} = 0,8mol\\
\dfrac{x}{y} = \dfrac{{0,3}}{{0,8}} = \dfrac{3}{8}\\
\Rightarrow CTDGN:{({C_3}{H_8})_n}\\
M = 22 \times {M_{{H_2}}} = 22 \times 2 = 44\\
12 \times 3n + 8n = 44\\
\Rightarrow n = \dfrac{{44}}{{44}} = 1\\
\Rightarrow CTPT:{C_3}{H_8}
\end{array}\)
