kwas siarkowy (VI) - H2SO4 m H2SO4= 2 * H + S+ 4 * 0 m H2SO4= 2* 1g + 32g + 4 * 16g= 98g m H = 1 u= 1g m S= 32 u=32g m O= 16u=16g % m H= 98 g - 100 % 2 g - x x = 2* 100 % / 98 % x = 2,04 % % m S= 98 g - 100 % 32 g - x x = 32 * 100% / 98 x = 32,65 % % m O = 98 g - 100% 64 g- x x= 64 * 100%/ 98 x = 65,3 % kwas siarkowy (IV) - H2SO3 m H2SO3 = 2 * m H + m S + 3 * m O = 2 * 1 g + 32 g + 3 * 16 g = 82 g m H = 1 g m S = 32 g m O = 16 g % m H = 82 g - 100 % 2 g - x x = 2* 100%/ 82 x = 2,43 % % m S = 82 g - 100 % 32 g - x x= 32 * 100%/ 82 x = 39,02 % % m O = 82 g - 100 % 48 g - x x = 48 * 100%/ 82 x = 58,53 % proszę ;DD
