`nAgCl=0,3 (mol)`
`ACl + AgNO3 -> ANO3 + AgCl`
` a.......................................a`
`BCl + AgNO3 -> BNO3 + AgCl`
` b.......................................b`
`a, => n X = nAgCl=0,3 (mol)`
` =>` `overline{X}` `=19,15:0,3=63,833`
`=> A <` `overline{X}` ` < B`
` => NaCl(58,5) < 63,833 < KCl(74,5)`
`b,` Ta có `:`
$\left \{ {{58,5a+74,5b=19,15} \atop {a+b=0,3}} \right.$
` => a=0,2 ; b=0,1 `
` m dd mới = mX + mddAgNO3 - mAgCl`
` = 19,15 + 300- 43,05 =276,1 (g)`
`m NaCl = 0,2.58,5=11,7 (g)`
` => % NaCl = 11,7:276,1.100=4,23%`
`m KCl = 0,1.74,5=7,45 (g)`
` => %KCl =7,45:276,1.100=2,69%`