`nHCl = 0.25 * 1.5 = 0.375 ( mol) `
`KOH + HCl -> KCl + H2O`
`0.375 - 0.375 - 0.375`
`a) V_(KOH) = 0.375 / 2 = 0.1875 ( l )`
`b) Cm_(KCl) =` $\frac{0.375}{0.1875 + 0.25 }$ `= 0.86 M`
Câu 2:
`n_(FeCl_2) = 0.2 * 0.15 = 0.03 ( mol)`
`FeCl2 + 2NaOH -> 2NaCl + Fe(OH)2↓`
` 0.03 - 0.06 - 0.03`
`4Fe(OH)2 + O2 -> 2Fe2O3 + 4H2O`
` 0.03 0.015`
`b) m_(Fe2O3) = 0.015 * 160 = 2.4 (g)`
`c) Cm_(NaCl) =` $\frac{0.06}{0.2 + 0.3}$ ` = 0.12 M`