$\text{a) Ta có: p + e + n = 49}$
$\text{Mà do: p = e} \to \text{p + e = 2p}$
$\to \text{ 2p + n = 49}$
$\text{Ta lại có: n=53,125%2p}=\dfrac{53,125}{100}2p=\text{0,53125.2p}$
$\to \text{2p +n =49} \to \text{2p= 49-n (1)}$
$\text{Do n=0,53125.2p, thế (1) vào ta có:}$
$\text{ n=0,53125(49-n)}$
$\to \text{ n=26,03125-0,53125n}\to n+0,53125n=26,03125$
$\to n=\dfrac{26,03125}{1,53125}=17$
$\to \text{2p= 49-17=32}$
$\to \text{p=e=}\dfrac{32}{2}=16$
$\to \text{R là S(p=16)}$
$\text{b) CTHH của R(II) và H:}$
$\text{Do ta có R là S (câu a)}$
$\text{Cho công thức có dạng:} H_xS_y$
$\to \text{ Theo quy tắc hoá trị, ta có:}$
$1x=2y$
$\to \dfrac{x}{y}=\dfrac{2}{1}$
$\to x=2, y=1$
$\text{Vậy CTHH là:} H_2S$
$\text{c)}$
$\text{Phương trình:}$
$ S + O_2 \xrightarrow{t^0} SO_2$
$\text{Ta có:}$
$\overline{M}_{hh\text{..khí}}=25,6M_{H_2}$
$\to \overline{M}_{hh\text{..khí}}=51,2\text{(g/mol)}$
$\text{mà:} \overline{M}_{hh\text{..khí}}=\dfrac{m_1+m_2}{n_1+n_2}=51,2$
$\to \overline{M}_{hh\text{..khí}}=\dfrac{m_1+m_2}{\dfrac{m_1}{32}+\dfrac{m_2-m_1}{32}}=51,2$
$\to \overline{M}_{hh\text{..khí}}=\dfrac{m_1+m_2}{\dfrac{m_1+m_2-m_1}{32}}=51,2$
$\to \overline{M}_{hh\text{..khí}}=\dfrac{m_1+m_2}{\dfrac{m_2}{32}}=51,2$
$\to \dfrac{m_1+m_2}{m_2}{m_2}=1,6$
$\to m_1+m_2=1,6m_2$
$\to m_1=0,6m_2$
$\to \dfrac{m_1}{m_2}=\dfrac{0,6}{1}=\dfrac{3}{5}$
