$\text{a) Ta có: p + e + n = 49}$

$\text{Mà do: p = e} \to \text{p + e = 2p}$

$\to \text{ 2p + n = 49}$

$\text{Ta lại có: n=53,125%2p}=\dfrac{53,125}{100}2p=\text{0,53125.2p}$

$\to \text{2p +n =49} \to \text{2p= 49-n (1)}$

$\text{Do n=0,53125.2p, thế (1) vào ta có:}$

$\text{ n=0,53125(49-n)}$

$\to \text{ n=26,03125-0,53125n}\to n+0,53125n=26,03125$

$\to n=\dfrac{26,03125}{1,53125}=17$

$\to \text{2p= 49-17=32}$

$\to \text{p=e=}\dfrac{32}{2}=16$

$\to \text{R là S(p=16)}$

$\text{b) CTHH của R(II) và H:}$

$\text{Do ta có R là S (câu a)}$

$\text{Cho công thức có dạng:} H_xS_y$

$\to \text{ Theo quy tắc hoá trị, ta có:}$

$1x=2y$

$\to \dfrac{x}{y}=\dfrac{2}{1}$

$\to x=2, y=1$

$\text{Vậy CTHH là:} H_2S$

$\text{c)}$

$\text{Phương trình:}$

$ S + O_2 \xrightarrow{t^0} SO_2$

$\text{Ta có:}$

$\overline{M}_{hh\text{..khí}}=25,6M_{H_2}$

$\to \overline{M}_{hh\text{..khí}}=51,2\text{(g/mol)}$

$\text{mà:} \overline{M}_{hh\text{..khí}}=\dfrac{m_1+m_2}{n_1+n_2}=51,2$

$\to \overline{M}_{hh\text{..khí}}=\dfrac{m_1+m_2}{\dfrac{m_1}{32}+\dfrac{m_2-m_1}{32}}=51,2$

$\to \overline{M}_{hh\text{..khí}}=\dfrac{m_1+m_2}{\dfrac{m_1+m_2-m_1}{32}}=51,2$

$\to \overline{M}_{hh\text{..khí}}=\dfrac{m_1+m_2}{\dfrac{m_2}{32}}=51,2$

$\to \dfrac{m_1+m_2}{m_2}{m_2}=1,6$

$\to m_1+m_2=1,6m_2$

$\to m_1=0,6m_2$

$\to \dfrac{m_1}{m_2}=\dfrac{0,6}{1}=\dfrac{3}{5}$

 

a, Ta có: $p+e+n=49$⇒$2p+n=49$

Mà $n=0,53125.2p$

⇒$3,0625p=49$

⇒$p=16$

⇒ R là S (lưu huỳnh)

b, Đătk CTHH dạng $H_xS_y$

Áp dụng quy tắc đương chéo , ta có:

$x.I=y.II$

⇒$x:y=2:1$⇒$x=2; y=1$

⇒CTHH dạng $H_2S$

c, $S+O_2\buildrel{{t^o}}\over\to SO_2$

Ta có: $\dfrac{m_1+m_2}{\dfrac{m_1}{32}+\dfrac{m_2-m_1}{32}}=51,2$  

⇒$\dfrac{m_1+m_2}{\dfrac{m_2}{32}}=51,2$  

⇒$\dfrac{m_1+m_2}{m_2}=1,6$  

⇒$m_1+m_2=1,6m_2$

⇒$m_1=0,6m_2$

⇒$\dfrac{m_1}{m_2}=\dfrac{3}{5}$