On Mole Ratio

2019 O Level, Combined Chemistry, Open-ended Question 8c

In the Mond process, nickel reacts with carbon monoxide to give the gas nickel carbonyl, Ni(CO)4.

Ni(s) + 4CO(g) ⟶ Ni(CO)4(g)

(a) Calculate the mass of nickel needed to produce 1000 g of nickel carbonyl.
(b) Calculate the volume of carbon monoxide at room temperature and pressure that is needed to react with this mass of nickel.

Answer: 345 g of Ni, 561 dm3 of CO

STEP 1: Convert data given to number of moles
Molar mass of Ni(CO)4 = 59 + 4×12 + 4×16 = 171 g/mol
No of moles of Ni(CO)4 = 1000 ÷ 171 = 5.848 mol

STEP 2: State the mole ratio of Ni(CO)4 to Ni
Mole ratio of Ni(CO)4 to Ni = 1:1

STEP 3: Use the mole ratio to find the amount of Ni needed
No of moles of Ni needed = (1/1) × 5.848 = 5.848 mol

STEP 4: Find the mass of Ni needed, as required by the question
Molar mass of Ni = 59 g/mol
Mass of Ni needed = 5.848 × 59 = 345 g (3 s.f.)


STEP 1: State the mole ratio of CO to Ni
Mole ratio of CO to Ni = 4:1

STEP 2: Use the mole ratio to find the amount of CO needed
No of moles of CO needed to react with 5.848 mol of Ni = (4/1) × 5.858 = 23.39 mol

STEP 3: Find the volume of CO needed, as required by the question
Gaseous volume of 23.39 mol of CO = 24 × 23.39 = 561 dm3 (3 s.f.)

On Concentration and Titration

2019 O Level, Combined Chemistry, MCQ 7

0.1 mol/dm3 hydrochloric acid reacts with 25.0 cm3 of 0.2 mol/dm3 aqueous sodium carbonate. The equation for this reaction is: 2HCl + Na2CO3 ⟶ 2NaCl + H2O + CO2

What is the volume of acid required to completely react with this volume of sodium carbonate?

A. 6.25 cm3
B. 25 cm3
C. 50 cm3
D. 100 cm3

Answer: 100 cm3

Method 1: By Calculation

No of moles of Na2CO3 = 0.2 × (25/1000) = 0.005 mol

Mole ratio of HCl to Na2CO3 = 2:1
No of moles of HCl = 0.005 × (2/1) = 0.01 mol

Volume of HCl = 0.01 ÷ 0.1 = 0.1 dm3 = 100 cm3

Method 2: By Reasoning

Noticeably, the concentration of hydrochloric acid is half of sodium carbonate. To keep the amount of solutes constant, we must make it up by doubling the volume. However, from the mole ratio, we need two times the amount of hydrochloric acid than there is sodium carbonate. Therefore, the volume must be doubled again. This means that we need 25 cm3 × 2 × 2 = 100 cm3 of hydrochloric acid.

2019 O Level, Combined Chemistry, Open-ended Question 9c

Sodium sulfate is a soluble salt. To prepare sodium sulfate, sodium hydroxide is added to sulfuric acid.

(a) The molar concentration of a solution of sodium hydroxide is 3 mol/dm3. Calculate the mass of sodium hydroxide required to form 4 dm3 of this solution.
(b) Write the chemical equation for the reaction between sodium hydroxide and sulfuric acid.
(c) Using the chemical equation, calculate the number of moles of acid needed to neutralise the solution of sodium hydroxide made in (a).

STEP 1: Convert data given to number of moles
No of moles of NaOH = 3 × 4 = 12 mol

STEP 2: Find the mass using molar mass
Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol
Mass of NaOH = 12 × 40 = 480 g


2NaOH + H2SO4 ⟶ Na2SO4 + 2H2O


STEP 1: State the mole ratio of H2SO4 to NaOH
Mole ratio of H2SO4 to NaOH = 1:2

STEP 2: Using the mole ratio, find the number of moles of H2SO4 to neutralise 12 mol of NaOH
No of moles of H2SO4 = (1/2) × 12 = 6.00 mol

2018 O Level, Combined Chemistry, MCQ 6

Acid rain contains sulfuric acid. 25 cm3 of acid rain is neutralised by 28 cm3 of 0.5 mol/dm3 aqueous sodium hydroxide: H2SO4 + 2NaOH ⟶ Na2SO4 + 2H2O

What is the concentration of sulfuric acid in the acid rain?

A. 0.28 mol/dm3
B. 0.56 mol/dm3
C. 1.12 mol/dm3
D. 28.00 mol/dm3

Answer: A

STEP 1: Convert data given to number of moles
No of moles of NaOH = 0.5 × (28/1000) = 0.014 mol

STEP 2: State the mole ratio of H2SO4 to NaOH
Mole ratio of H2SO4 to NaOH = 1:2

STEP 3: Use the mole ratio to find the number of moles of H2SO4
No of moles of H2SO4 = (1/2) × 0.014 = 0.007 mol

STEP 4: Find the concentration of H2SO4
Concentration of H2SO4 = no of moles ÷ volume = 0.007 ÷ (25/1000) = 0.28 mol/dm3

2018 O Level, Combined Chemistry, Open-ended Question 8

Some indigestion tablets reduce stomach acidity. Carbonate ions, CO32-, in these tablets react with hydrogen ions in the stomach to form water and a gas.

(a) Write the ionic equation for this chemical reaction.
(b) Calculate how many moles of hydrogen ions are present in 500 cm3 of a 0.08 mol/dm3 solution of hydrogen ions.
(c) Given that 100 indigestion tablets contain 1 mol of carbonate ions. Calculate how many indigestion tablets are needed to neutralise the acidic solution.

2H+(aq) + CO32-(aq)⟶ H2O(l) + CO2(g)


No of moles of H+ ions = 0.08 × (500/1000) = 0.0400 mol


STEP 1: State the mole ratio of CO32- to H+
Mole ratio of CO32- to H+ = 1:2

STEP 2: Use the mole ratio to find the number of moles of CO32-
No of moles of CO32- = (1/2) × 0.04 = 0.02 mol


No of digestion tablets needed = 0.02 mol × 100 tablets/mol = 2 tablets