Calculate the volume of Ba(OH)2 which would be required to react with 25 cm3 of 12g of CH3COOH?

Question

Ba(OH)2 + 2CH3COOH ----> (CH3COO)2 Ba
0.1 M 12 g
25 cm3

0.1 M of Ba(OH)2

Answer 1

Ba(OH)2 + 2CH3COOH ----> (CH3COO)2 Ba

mole CH3COOH: 12g/ 60g/mol = 0.2 mol
mole Ba(OH)2: 0.2 mol/ 2 = 0.1 mol
- since 2 mol acetic acid is required for every mole Ba(OH)2

If concentration of Ba(OH)2 is 0.1M, volume required to react is 0.1 mol/ 0.1 M = 1L or 1000 cm3