Дано:
$H_{2}SO_{4}$
$H_{2}SO_{3}$
$H_{2}S$
Найти:
$ω_{H_{2}SO_{4}}$ ($S$) − ?
$ω_{H_{2}SO_{3}}$ ($S$) − ?
$ω_{H_{2}S}$ ($S$) − ?
Решение:
$M_{r}$ ($H_{2}SO_{4}$) = 2 * 1 + 32 + 4 * 16 = 98
$M_{r}$ ($H_{2}SO_{3}$) = 2 * 1 + 32 + 3 * 16 = 82
$M_{r}$ ($H_{2}S$) = 2 * 1 + 32 = 34
ω = $\frac{n\; * \;A_{r}}{M_{r}}$ * 100%
$ω_{H_{2}SO_{4}}$ ($S$) = $\frac{1\; * \;32}{98}$ * 100% = 32,65%
$ω_{H_{2}SO_{3}}$ ($S$) = $\frac{1\; * \;32}{82}$ * 100% = 39%
$ω_{H_{2}S}$ ($S$) = $\frac{1\; * \;32}{34}$ * 100% = 94,1%
$ω_{H_{2}SO_{4}}$ ($S$) < $ω_{H_{2}SO_{3}}$ ($S$) < $ω_{H_{2}S}$ ($S$)
Ответ: $ω_{H_{2}SO_{4}}$ ($S$) = 32,65%; $ω_{H_{2}SO_{3}}$ ($S$) = 39%; $ω_{H_{2}S}$ ($S$) = 94,1%.