Form 5 Chemistry Practice Paper: Rates Of Reaction, Titration And Redox (12 Questions, 100 Marks, 75 Minutes)

Form 5 Chemistry – 12 questions – 100 marks – 75 minutes

Form 5 Chemistry practice paper: 12 questions, 100 marks, 75 minutes on reaction rates, titration, dilution, pH and redox, with full worked answers.

Paper details　Form 5 Chemistry practice paper with 12 original questions on reaction rates, titration and redox, plus full worked answers.

Section A: Rates of reaction (Q1-Q4)

Show the working steps, then the units and the final answer.

#	Question	Answer and marking notes
1	25.0 cm3 of 2.00 mol/dm3 hydrochloric acid is added to excess magnesium ribbon. Calculate the number of moles of hydrogen ions available.	Answer: 0.0500 mol. Moles of hydrogen ions = 0.0250 dm3 x 2.00 mol/dm3 = 0.0500 mol.
2	The reaction in Q1 collects 96 cm3 of hydrogen in 40 seconds at room temperature. Calculate the average rate of hydrogen formation in cm3/s.	Answer: 2.4 cm3/s. Average rate = 96 cm3 / 40 s = 2.4 cm3/s.
3	In a reaction the mass of a reactant falls from 3.00 g to 1.80 g in 30 seconds. Calculate the average rate of loss of mass in g/s.	Answer: 0.0400 g/s. Mass lost = 3.00 g – 1.80 g = 1.20 g; rate = 1.20 g / 30 s = 0.0400 g/s.
4	Ammonia is synthesised by N2 + 3H2 -> 2NH3. If nitrogen is consumed at 0.020 mol/s, find the rate of consumption of hydrogen and the rate of formation of ammonia.	Answer: hydrogen 0.060 mol/s and ammonia 0.040 mol/s. From the ratio 1:3:2, hydrogen = 3 x 0.020 = 0.060 mol/s and ammonia = 2 x 0.020 = 0.040 mol/s.

For acid-base questions write the equation and the mole ratio.

#	Question	Answer and marking notes
5	25.0 cm3 of 0.100 mol/dm3 sodium hydroxide is neutralised by 0.050 mol/dm3 sulphuric acid. Calculate the volume of acid needed in cm3.	Answer: 25.0 cm3. Moles of sodium hydroxide = 0.100 x 0.0250 = 0.00250 mol; moles of acid = 0.00250 / 2 = 0.00125 mol; volume = 0.00125 / 0.050 x 1000 = 25.0 cm3.
6	A solution has a hydrogen ion concentration of 1.0 x 10-3 mol/dm3 (Kw = 1.0 x 10-14). Find its pH and its hydroxide ion concentration.	Answer: pH 3.00 and hydroxide concentration 1.0 x 10-11 mol/dm3. pH = -log(1.0 x 10-3) = 3.00; [OH-] = 1.0 x 10-14 / 1.0 x 10-3 = 1.0 x 10-11.
7	10.0 cm3 of 2.00 mol/dm3 hydrochloric acid is diluted with water to 250 cm3. Calculate the new concentration in mol/dm3.	Answer: 0.0800 mol/dm3. Moles before dilution = 2.00 x 0.0100 = 0.0200 mol; concentration = 0.0200 / 0.250 = 0.0800 mol/dm3.
8	5.00 g of limestone reacts completely with 0.0900 mol of hydrochloric acid: CaCO3 + 2HCl -> CaCl2 + CO2 + H2O. Calculate the mass of calcium carbonate in the sample and its percentage by mass.	Answer: 4.50 g and 90.1 per cent. Moles of calcium carbonate = 0.0900 / 2 = 0.0450 mol; mass = 0.0450 x 100.09 = 4.50 g; percentage = 4.50 / 5.00 x 100 = 90.1 per cent.

Show the oxidation number calculation; give electrons in moles.

#	Question	Answer and marking notes
9	State the oxidation number of manganese in KMnO4 and of chromium in K2Cr2O7.	Answer: manganese +7 and chromium +6. For KMnO4, 1 + x + 4(-2) = 0 so x = +7; for K2Cr2O7, 2(+1) + 2x + 7(-2) = 0 so x = +6.
10	0.0200 mol of potassium manganate(VII) is reduced to manganese(II) ions in acid solution. Each manganate(VII) ion accepts 5 electrons; calculate the number of moles of electrons transferred.	Answer: 0.100 mol. Moles of electrons = 0.0200 mol x 5 = 0.100 mol.
11	The electrons from Q10 oxidise iron(II) ions to iron(III) ions. Calculate the mass of iron oxidised in grams (Fe = 55.85).	Answer: 5.59 g. Each iron(II) ion loses one electron, so moles of iron = 0.100 mol; mass = 0.100 x 55.85 = 5.59 g.
12	What is the volume of 0.0500 mol of hydrogen gas at room temperature in dm3? (molar volume = 24.0 dm3/mol)	Answer: 1.20 dm3. Volume = 0.0500 mol x 24.0 dm3/mol = 1.20 dm3.

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