xmol 0.0054mol
Ba(OH)2 + 2HBr = BaBr2 +2H2O
1mol 2mol
n(HBr) = C(HBr) × V(HBr) = 0.225M × 0.024L = 0.0054mol.
n(Ba(OH)2) = x = 0.0054 × 1 / 2 mol = 0.0027mol.
V(Ba(OH)2) = n(Ba(OH)2) / C(Ba(OH)2) = 0.0027mol / 0.171M = 0.0158L = 15.8mL.
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