Đáp án:

\(8.\, C_2H_6\\ 9. \, a)\, Fe_2O_3\\ b)\, CO\\ 10. \, N_2O_5\)

Giải thích các bước giải:

\(8.\,B: C_xH_y\\d_{B/H_2}=15\to M_B=15\times 2=30\text{(g/mol)}\to 12x+y=30\\ \dfrac{\%m_C}{\%m_H}=\dfrac{80\%}{20\%}\to \dfrac {12x}y=4\to 12x-4y=0\to \begin{cases}x=2\\ y=6\end{cases} \to C_2H_6\)

\(9.\, a)\, CTHH: Fe_xO_y\\\to \dfrac{56x}{16y}=\dfrac 73\to \dfrac xy=\dfrac 23\to Fe_2O_3\\ b)\, A:\, C_xH_y\\\to x:y=\dfrac{12}3:\dfrac 41=4:4=1:1\to CO \)

\(10. \, CTHH: N_xO_y\\\to \dfrac{14x}{16y}=\dfrac 7{20}\to \dfrac xy=\dfrac 25\to N_2O_5\)

Đáp án:

 8) C2H6

9) a) Fe2O3

     b) CO

10) N2O5

Giải thích các bước giải:

\(\begin{array}{l}
8)\\
CTHH:{C_x}{H_y}\\
x:y = \dfrac{{80}}{{12}}:\dfrac{{20}}{1} = 1:3\\
 \Rightarrow CTDGN:C{H_3}\\
B:{(C{H_3})_n}\\
MB = 15 \times 2 = 30g/mol\\
 \Rightarrow 15n = 30 \Rightarrow n = 2\\
 \Rightarrow CTPT:{C_2}{H_6}\\
9)\\
a)\\
CTHH:F{e_x}{O_y}\\
x:y = \dfrac{7}{{56}}:\dfrac{3}{{16}} = 2:3\\
 \Rightarrow CTHH:F{e_2}{O_3}\\
b)\\
CTHH:{C_x}{O_y}\\
x:y = \dfrac{3}{{12}}:\dfrac{4}{{16}} = 1:1\\
 \Rightarrow CTHH:CO\\
10)\\
CTHH:{N_x}{O_y}\\
x:y = \dfrac{7}{{14}}:\dfrac{{20}}{{16}} = 2:5\\
 \Rightarrow CTHH:{N_2}{O_5}
\end{array}\)