nNaOH =0,3 mol
Ưu tiên phản ứng giữa axit bazo phản ứng trước:
......__HCl + NaOH → NaCl + H2O
bd: __0,05......0,3...........0..........
pứ: __0,05--->0,05--->0,05---->0,05 ( Vì \(\frac{n_{HCl}}{1}< \frac{n_{NaOH}}{1}\)=> HCl hết NaOH dư=> Tính theo HCl)
sau:___0_......0,25.......0,05.......
...........ZnSO4 + 2NaOH→ Zn(OH)2 + Na2SO4
bd: ___0,08_____0,25______0_________0
pứ: ___0,08------->0,16------>0,08-------->0,08 ( Vì \(\frac{n_{ZnSO4}}{1}< \frac{n_{NaOH}}{2}\)=> ZnSO4 hết NaOH dư=> Tính theo ZnSO4)
sau: ___0_______0,09_____0,08_______0,08
_____Zn(OH)2 + 2NaOH → Na2ZnO2 + 2H2O
bd: ___0,08_____0,09_________0_
pư: ___0,045<-----0,09____ (Vì \(\frac{n_{Zn\left(OH\right)_2}}{1}< \frac{n_{NaOH}}{2}\)=> NaOH hết, Zn(OH)2 dư=> Tính theo NaOH
sau:___0,035______0_____________
=> nZn(OH)2=0,035 mol
=>m↓=mZn(OH)2=0,035*99=3,465 (g)
