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

i know 3 people have already tried to answer this question for me but i still dont understand....

Answer 1

Hi! I'm brazillian and i don't know english very well but i'll try to answer you, ok?

Organization

HCl Ba(OH)

M= 2mol/dm3 M= ? *
V=? ** V= 50


To completely react:

n° mol [H]+ = n° mol [OH]-

[M x X x V] acid = [M x X x V] hydroxide

2 x 1x V = M x 2 x 50

Lets calculate mol:

1mol Ba(OH)2 -------------- 171.5 g
x mol --------------- 68.4 g


mol = 0.3988..... ~= 0.4 mol *

Now you substitute:

2 x 1 x V = 0.4 x 2 x 50

2V = 40 ====> R: 20 cm3 **


Bye!

Answer 2

First calculate the number of moles in 50 cm3 of Ba(OH)2.

50 cm2 x 1dm3/1000cm3 x 68.3g/1 dm3 x 1mol/171 g = 0.02 moles

Next calculate the concentration of HCl in mols per cm.
2.0 mol/1dm3 x 1dm3/1000 cm3 = 2E-3 mol/cm3

Then calc the the volume of HCl needed by using the molar relationship in the balanced equation.

0.02 mol Ba(OH)2 x 2 mol HCl/1mol Ba(OH)2 x 1 cm3/2E-3 mol HCl. = 20.0 cm3

NOTE 2 E-3 = 2 x 10^-3

Answer 3

The Ba(OH)2 solution contains:

m = 0.05 x 68.4 = 3.42 g of Ba(OH)2

The relative molar mass Mr of Ba(OH)2 equals 171 g/mol (approx), so the quantity of Ba(OH)2 in moles is:

n = m/Mr, n = 3.42/171 = 0.02 mol

According to the chem. equation

1 mol of Ba(OH)2 reacts with 2 mol of HCl, so
0.02 mol of Ba(OH)2 reacts with 0.04 mol of HCl

Now we can find the volume of the HCl solution:

n = C*V, so V = n/C where n = 0.04 mol the moles of HCl and C = 2 M the concentration of the solution

V = 0.04/2 = 0.02 L or 20 mL (20 cm3)

Answer 4

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