Calculations on the mole
Francine Taylor-Campbell, Contributor

Students of Edith Dalton-James High School go wild at the STAR's 'No Violence in Schools Tour' concert, December 5, 2006. - Ian Allen/Staff Photographer

Let's consider the following question. 1. 34 cm³ of hydrochloric acid of unknown concentration required 25 cm³ NaOH of concentration 2.00 mol/dm³ to be completely neutralised.

(a) Write the equation for the reaction between sodium hydroxide and hydrochloric acid. (Include state symbols)

(b) Calculate the number of moles of sodium hydroxide in 25 cm³ of the solution used.

(c) Calculate the number of moles of hydrochloric acid in the volume of hydrochloric acid used.

(d) Calculate the number of moles of hydrochloric acid in 1 dm³ of solution.

ANSWER

(a) HCl (aq) + NaOH (aq) == NaCl (aq) + H2O (l).

(b) Concentration of NaOH is 2.0 mol/dm³

Thus 2 mol NaOH are present in 1000 cm3

X mol NaOH is present in 25 cm³.

X mol = (25 x 2)/1000 = 0.050 mol

(c) Since NaOH and HCl react in a 1:1 ratio, the number of moles of HCl that reacted is also 0.050 mol which is present in 34 cm³.

(d) 0.050 mol HCl is in 34 cm³.

Thus X mol are in 1000 cm3 (1 dm³)

X mol = (0.050 x 1000)/34 = 1.47 mol

Concentration of HCl = 1.47 mol/dm³.

Question 2

28.50 cm³ of 0.050 mol/dm³ H2SO4 exactly neutralised 25.00cm³ X2CO3 of concentration 6.00g/dm3. Calculate (a) Mr for X2CO3 (b) Ar for X:

(i) Equation: X2CO3 + H2SO4 = X2SO4 + H2O + CO2

28.50cm3 H2SO4 contains the same number of moles as 25.00cm3 X2CO3.

(ii) moles of H2SO4 in 28.50cm3 of 0.050 mol/dm³ = (28.50x0.050)/1000 = 0.001425 mols.

(iii) moles X2CO3 in 25.00cm³ = 0.001425 mols.

(iv) Moles X2CO3 in 1000cm³ = (0.001425x1000)/25 = 0.057 mol/dm³.

(v) But 1dm³ X2CO3(aq) contains 6.00g

6.00g has 0.057 mol

Therefore 6.00/0.057 = 1 mol = 105g

Mr = 105.0.

(vi) Mr of X2CO3 = (Ar of Xx2) + 12 + (3x16) = 105

Ar of X = (105 ñ (12 48))2 = 22.5 i Ar = 22.5.

Question 3

A small piece of lithium of mass 0.35g is added to cold water. The resulting solution is titrated with 2.00 mol/dm³ hydrochloric acid.

(a) Write the equation between lithium and water and the lithium solution and hydrochloric acid.

(b) What volume of hydrochloric acid is needed to neutralise the lithium solution?

ANSWERS

a) 2Li (s) + 2H²O (l) == 2LiOH (aq) + H2 (g)

Remember that the alkali metals dissolve in water to form a metal hydroxide and give off hydrogen.

LiOH (aq) + HCl (aq) == LiCl (aq) + H2O (l)

(b) Molar mass of lithium is 7

# mol Li = 0.35/7 = 0.05 mol

Based on the equation Li reacts in a 1:1 ratio to form LiOH

Thus the # mol LiOH = 0.05 mol

LiOH and HCl also react in a 1:1 ratio

# mol of HCl is also 0.05 mol

Since 1000 cm³ contain 2.00 mol

X cm³ contain 0.05 mol thus X cm³ = (1000 x 0.05)/2 = 25 cm³.

Now attempt these questions:

a) What is the concentration of a solution of sodium hydroxide, if 25cm³ of it requires 20cm³ of hydrochloric acid of concentration 0.100 mol/dm3 for neutralisation?

b) 37.50cm³ of HCl containing 0.100 mol/dm³ neutralised 25.00cm³ XHCO3 of concentration 15.00g/dm³. Calculate Mr for XHCO³ and Ar for X.

c) In this titration 25.0 cm³ 0f 1.0 mol/dm³ NaOH was used. Calculate the volume of 2.0 mol/dm³ HCl needed to neutralise the alkali. Calculate the mass of sodium chloride formed.

Francine Taylor-Campbell is an independent contributor.