... l m KCl.r mKC1,B| %m = 0,08 : 0 ,12 : 0,05 = : 12 : V a y c o n g thufc ciia c h a t h a u co A l a C8H12O5 c6 M A < B 56,72% KCl (pt4) , = 8,94 22 29,8 74,5 122 ,5 = 49 g a m 49.100 KC10.,(A) ... muoi c6 so m o l b^ng A B C Hiidng va bang x 136x + 123 x + 190x = 8,98 =>x = 0,02; dan D gidi n„^o, = 0.25 0,5 = 0 ,125 m o l => n,j = 0 ,125 = 0,25 m o l nHci = 0,25 = 0,25 m o l => n „ = ... phuong t r i n h (1),(2) t a t h a y •^muo'i clorua ~ ^ Pt: H2O 197 = 0,12mol -'^cOo B2CO3 + H2SO4 ^ | y = 0 ,12 "co, =0 ,12 + 0,26 = 0,38 mol (loang) ^ (loang) ^ A S O + CO2T + H O A S O + CO2T...