Calculations on the mole
Francine Taylor-Campbell, Contributor

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.

ANSWERS

(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 1,000 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

2. 28.50 cm³ of 0.050 mol/dm³ H₂SO₄ exactly neutralised 25.00cm³

X2CO³ of concentration 6.00g/dm³. Calculate (i) Mr for X₂CO₃ (ii) 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

ie 6.00/0.057 = 1 mol = 105g

Mr = 105.0

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

Ar of X = (105 - (12 + 48))/2 = 22.5 ie Ar = 22.5

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 neutralize 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

Q3. 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 neutralization?

Q4. 37.50cm³ of HCl containing 0.100 mol/dm³ neutralized 25.00cm³ XHCO₃ of concentration 15.00g/dm³. Calculate Mr for XHCO₃ and Ar for X

Q5. 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 neutralize the alkali. Calculate the mass of sodium chloride formed.

Francine Taylor-Campbell is an independent contributor. Send questions and comments to kerry-ann.hepburn@gleanerjm.com