Calcualting the colume of acid; include all steps involved in working out the answer please?

Question

how many cubic centimetres of a solution of hydrochloric acic, of concentration 2 mol dm-3, are required to react completely with 50cm3 of a solution of barium hydroxide that contains 68.4 g dm-3?

2HCl (aq) + Ba(OH)2 (aq) ==> BaCl2 (aq) + 2H2O(l)

the answer given is actually 20 cm3, but i just dont know how to get to that, so can you show all the steps of your working out of how to get to that answer. thanks

Answer 1

Certain terminally silly chemistry teachers give concentrations in dm^-3, which is actually 1L = 1000mL = 1000 cu cm, and which is confusing to students who have enough trouble just learning chemistry.

Atomic weights: Ba = 137 O = 16 H = 1 Ba(OH)2 = 171

50cm3Ba(OH)2 x 68.4gBa(OH)2/1000cm3Ba(OH)2 x 1molBa(OH)2/171gBaOH)2 x 2molHCl/1molBa(OH)2 x 1000cm3HCl/2molHCl = (50)(68.4)(2)(1000)/(1000)(171)(2) = (50)(68.4)/(171) = 20cm3HCl

The 50cm3 Ba(OH)2 is given. The first factor comes from the given concentration of Ba(OH)2. The cm3 solution cancel, leaving gBa(OH)2. The second factor comes from the molecular weight of Ba(OH)2. The gBa(OH)2 cancel, leaving molBa(OH)2. The third factor comes from the balanced equation. The molBa(OH)2 cancel, leaving molHCl. The fourth factor comes from the given concentration of HCl. The molHCl cancel, leaving cm3HCl.

I rewrote all the numbers other than 1 in numerators and denominators to show how the two 1000's and the two 2's could cancel, leaving you with only three numbers to crunch at the end.

Answer 2

First.
Calculate the moles of barium hydroxide.
Using atomic masses and the number of atoms of element.

Ba x 1 = 1 x 137 = 137
O x 2 = 2 x 16 = 32
H x 2 = 2 x 1 = 2
Total 171 (Mr of Ba(OH)2
moles of Ba(OH)2 = 68.4 g / 171
- do - = 0.4 moles dm^-3

therefore moles of Ba(OH)2 in 50 cm^3 is
0.4 moles dm^-3 x 50 cm^3 / 1000cm^3
Moles Ba(OH)2 = 0.02
Moles HCl = 0.02 x 2 = 0.04 moles

volume (HCl) = 0.04 mol x 1000 cm^3 / 2 mol dm^3
= 20 cm^3