Coi n HCl = 1(mol) => m dd HCl = 146(gam)
KOH + HCl $\to$ KCl + H2O
n KCl = n KOH = n HCl = 1(mol)
=> m dd sau pư = 1.74,5/20% = 372,5(gam)
=> m dd KOH = 372,5 - 146 = 226,5(gam)
C% KOH = 1.56/226,5 .100% = 24,72%
=> a = 24,72
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