Q1.A saturated aqueous solution of magnesium hydroxide contains 1.17 × 10–3 g of Mg(OH)2 in 100 cm3 of solution. In this solution, the magnesium hydroxide is fully dissociated into ions.
What is the concentration of Mg2+(aq) ions in this solution?
A        2.82 × 10–2mol dm–3        
B        2.01 × 10–3mol dm–3        
C        2.82 × 10–3mol dm–3        
D        2.01 × 10–4mol dm–3        
(Total 1 mark)

Q2.A particular sample of iron ore contains 85% by mass of Fe2O3 (Mr= 159.6) and no other iron compound. The maximum mass of iron that could be extracted from 1.0 tonne of this ore is 
A       0.59 tonne
B       0.66 tonne
C       0.75 tonne
C       0.85 tonne
(Total 1 mark)

Q3.The removal of silicon dioxide with limestone in the Blast Furnace can be represented by the following equation.
CaCO3(s) + SiO2(s) → CaSiO3(l) + CO2(g)
The minimum mass of calcium carbonate needed to remove 1.00 tonne (1000 kg) of silicon dioxide is
A       0.46 tonne
B       0.60 tonne
C       1.67 tonne
D       2.18 tonne
(Total 1 mark)

Q4.Magnesium reacts with hydrochloric acid according to the following equation.
Mg + 2HCl  MgCl2 + H2
A student calculated the minimum volume of 2.56 mol dm–3 hydrochloric acid required to react with an excess of magnesium to form 5.46 g of magnesium chloride (Mr = 95.3).

Which of the following uses the correct standard form and the appropriate number of significant figures to give the correct result of the calculation?
A        4.476 × 10–2 dm3      
B        4.48 × 10–2 dm3       
C        4.50 × 10–2 dm3       
D        44.8 × 10–3 dm3       
(Total 1 mark)

Q5.An ester is hydrolysed as shown by the following equation.
RCOOR/ + H2O  RCOOH + R/OH
What is the percentage yield of RCOOH when 0.50 g of RCOOH (Mr = 100) is obtained from 1.0 g of RCOOR/ (Mr = 150)?
A            33%         
B            50%         
C            67%         
D            75%         
(Total 1 mark)

Q6.What is the total volume of gas remaining after 20 cm3 ethane are burned completely in 100 cm3 oxygen? All volumes are measured at the same pressure and the same temperature, which is above 100 °C.
C2H6   +   3O2    2CO2   +   3H2O
 
 	A	40 cm3		
 	B	100 cm3		
 	C	120 cm3		
 	D	130 cm3		

(Total 1 mark)

Q7.30 cm3 of xenon are mixed with 20 cm3 of fluorine. The gases react according to the following equation. Assume that the temperature and pressure remain constant.
Xe(g) + F2(g) → XeF2(g)
What is the final volume of gas after the reaction is complete?
 
 	A	50 cm3		
 	B	40 cm3		
 	C	30 cm3		
 	D	20 cm3		

(Total 1 mark)
Q8.Hydrolysis of the ester, CH3COOCH2CH2CH3, produces ethanoic acid. In an experiment, 2.04 g of the ester was used and 0.90 g of ethanoic acid was produced. The percentage yield of ethanoic acid was:
A       44
B       59
C       75
D       90
(Total 1 mark)

Q9.In a car airbag, sodium azide (NaN3) decomposes to form sodium metal and nitrogen gas.
2NaN3(s) → 2Na(s) + 3N2(g)
The sodium metal then reacts with potassium nitrate to produce more nitrogen gas.
10Na(s) + 2KNO3(s) → N2(g) + 5Na2O(s) + K2O(s)
If 2.00 mol of sodium azide react in this way, how many molecules of N2 will be formed?
(The Avogadro constant L = 6.022 × 1023 mol–1)
 
 	A	2.41 × 1024		
 	B	1.93 × 1024		
 	C	1.81 × 1024		
 	D	9.63 × 1023		

(Total 1 mark)

Q10.Propene can be made by the dehydration of propan-2-ol.
What is the percentage yield when 30 g of propene (Mr = 42.0) are formed from 50 g of propan-2-ol (Mr = 60.0)?
 
 	A	60%		
 	B	67%		
 	C	81%		
 	D	86%		

(Total 1 mark)
Q11.          A metal carbonate MCO3 reacts with hydrochloric acid as shown in the following equation.
MCO3  +  2HCl  →  MCl2  +  H2O  +  CO2
A 0.548 g sample of MCO3 reacted completely with 30.7 cm3 of 0.424 mol dm–3 hydrochloric acid.
(a)     (i)      Calculate the amount, in moles, of HCl which reacted with 0.548 g MCO3
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(1)
(ii)     Calculate the amount, in moles, of MCO3 in 0.548 g.
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(1)
(iii)     Calculate the relative formula mass of MCO3
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(1)
(b)     Use your answer from part (a)(iii) to deduce the relative atomic mass of metal M and suggest its identity.
(If you have been unable to calculate a value for the relative formula mass of MCO3 you should assume it to be 147.6 but this is not the correct answer.)
Relative atomic mass ..................................................................................
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Identity of M ..................................................................................................
(2)
Q12.          Under suitable conditions magnesium will react with dilute nitric acid according to the following equation.
Mg(s) + 2HNO3(aq) → Mg(NO3)2(aq) + H2(g)
A 0.0732 g sample of magnesium was added to 36.4 cm3 of 0.265 mol dm–3 nitric acid. The acid was in excess.
(a)     (i)      Calculate the amount, in moles, of magnesium in the 0.0732 g sample.
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(1)
(ii)     Hence calculate the amount, in moles, of nitric acid needed to react completely with this sample of magnesium.
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(1)
(iii)     Calculate the amount, in moles, of nitric acid originally added to this sample of magnesium.
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(1)
(iv)    Hence calculate the amount, in moles, of nitric acid that remains unreacted.
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(1)

(b)     In a second experiment, 0.512 mol of hydrogen gas was produced when another sample of magnesium reacted with dilute nitric acid. Calculate the volume that this gas would occupy at 298 K and 96 kPa. Include units in your final answer.
(The gas constant R = 8.31 J K–1 mol–1)
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(3)

(c)     Concentrated nitric acid reacts with magnesium to form an oxide of nitrogen which contains 30.4% by mass of nitrogen.
Calculate the empirical formula of this oxide of nitrogen. Show your working.
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(3)
(Total 10 marks)

Q13.          (a)     The mass of one mole of 1H atoms is 1.0078 g and that of one 1H atom is
1.6734 × 10–24 g.
Use these data to calculate a value for the Avogadro constant accurate to five significant figures. Show your working.
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(2)

(b)     How does the number of atoms in one mole of argon compare with the number of molecules in one mole of ammonia?
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(1)

(c)     A sample of ammonia gas occupied a volume of 0.0352 m3 at 298 K and 98.0 kPa.
Calculate the number of moles of ammonia in the sample.
(The gas constant R = 8.31 J K–1 mol–1)
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(3)

(d)     A solution containing 0.732 mol of ammonia was made up to 250 cm3 in a volumetric flask by adding water. Calculate the concentration of ammonia in this final solution and state the appropriate units.
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(2)
(e)     A different solution of ammonia was reacted with sulphuric acid as shown in the equation below.
2NH3(aq)   +   H2SO4(aq)   →   (NH4)2SO4(aq)
In a titration, 25.0 cm3 of a 1.24 mol dm–3 solution of sulphuric acid required 30.8 cm3 of this ammonia solution for complete reaction.
(i)      Calculate the concentration of ammonia in this solution.
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(ii)     Calculate the mass of ammonium sulphate in the solution at the end of this titration.
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(6)
(f)      The reaction of magnesium nitride, Mg3N2, with water produces ammonia and magnesium hydroxide. Write an equation for this reaction.
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(2)
(Total 16 marks)

Q14.          Ammonium sulfate reacts with sodium hydroxide to form ammonia, sodium sulfate and water as shown in the equation below.
(NH4)2SO4(s) + 2NaOH(aq) → 2NH3(g) + Na2SO4(aq) + 2H2O(l)
(a)	A 3.14 g sample of ammonium sulfate reacted completely with 39.30 cm3 of a sodium hydroxide solution.

(i)      Calculate the amount, in moles, of (NH4)2SO4 in 3.14 g of ammonium sulfate.
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(2)
(ii)     Hence calculate the amount, in moles, of sodium hydroxide which reacted.
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(1)
(iii)     Calculate the concentration, in mol dm–3, of the sodium hydroxide solution used.
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(1)
(b)     Calculate the percentage atom economy for the production of ammonia in the reaction between ammonium sulfate and sodium hydroxide.
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(2)
(c)     Ammonia is manufactured by the Haber Process.
N2 + 3H2  2NH3
Calculate the percentage atom economy for the production of ammonia in this process.
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(1)
(d)     A sample of ammonia gas occupied a volume of 1.53 × 10–2 m3 at 37 °C and a pressure of 100 kPa.
(The gas constant R = 8.31 J K–1 mol–1)
Calculate the amount, in moles, of ammonia in this sample.
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(3)

(e)     Glauber’s salt is a form of hydrated sodium sulfate that contains 44.1% by mass of sodium sulfate. Hydrated sodium sulfate can be represented by the formula Na2SO4.xH2O where x is an integer. Calculate the value of x.
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(3)
(Total 13 marks)
Q15.Hydrogen peroxide is sold commercially as an aqueous solution containing approximately 60 g dm–3 of hydrogen peroxide.
(a)     Use data from the Periodic Table to calculate the Mr of hydrogen peroxide. Give your answer to the appropriate precision.
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(1)
(b)     Calculate the concentration, in mol dm–3, of a solution containing 60.0 g dm–3 of hydrogen peroxide.
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(1)
(c)     The concentration of hydrogen peroxide in a hair bleach is 0.050 mol dm–3. Use your answer from (b) to calculate the dilution factor needed to make the commercial hydrogen peroxide solution suitable for use in this hair bleach. Show your working.
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(2)
(Total 4 marks)

Q16.(a)     Sodium hydrogencarbonate (NaHCO3) can also be used to neutralise ethanoic acid spillages. The equation for this reaction is shown below.
CH3COOH  +  NaHCO3   CH3COONa  +  H2O  +  CO2
State the ideal gas equation.
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(1)

(b)     There are several methods by which ethanoic acid is synthesised on an industrial scale. One method is the oxidation of butane in the presence of metal ion catalysts. 
Balance the equation given below which summarises this reaction.
..........C4H10  +  ..........O2   ..........CH3COOH  +  ..........H2O
(1)

(c)     A second method by which ethanoic acid is synthesised involves the oxidative fermentation of ethanol in the presence of bacteria. The equation representing this reaction is given below.
C2H5OH  +  O2   CH3COOH  +  H2O
In a small scale experiment using this second method it was found that 23.0 g of ethanol produced only 4.54 g of ethanoic acid. Calculate the percentage yield for this experiment.
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(2)
(Total 4 marks)

Q17.          In this question give all your answers to three significant figures.
Magnesium nitrate decomposes on heating to form magnesium oxide, nitrogen dioxide and oxygen as shown in the following equation.
2Mg(NO3)2(s) → 2MgO(s) + 4NO2(g) + O2(g)
(a)     Thermal decomposition of a sample of magnesium nitrate produced 0.741 g of magnesium oxide.
(i)      Calculate the amount, in moles, of MgO in 0.741 g of magnesium oxide.
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(2)
(ii)     Calculate the total amount, in moles, of gas produced from this sample of magnesium nitrate.
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(1)

(b)     In another experiment, a different sample of magnesium nitrate decomposed to produce 0.402 mol of gas. Calculate the volume, in dm3, that this gas would occupy at 333 K and 1.00 × 105 Pa.
(The gas constant R = 8.31 J K–1 mol–1)
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(3)
(c)     A 0.0152 mol sample of magnesium oxide, produced from the decomposition of magnesium nitrate, was reacted with hydrochloric acid.
MgO + 2HCl → MgCl2 + H2O
(i)      Calculate the amount, in moles, of HCl needed to react completely with the 0.0152 mol sample of magnesium oxide.
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(1)

(ii)     This 0.0152 mol sample of magnesium oxide required 32.4 cm3 of hydrochloric acid for complete reaction. Use this information and your answer to part (c) (i) to calculate the concentration, in mol dm–3, of the hydrochloric acid.
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(1)
(Total 8 marks)
Q18.Some airbags in cars contain sodium azide (NaN3).
(a)     Sodium azide is made by reacting dinitrogen monoxide gas with sodium amide (NaNH2) as shown by the equation.
2NaNH2  +  N2O    NaN3   +   NaOH   +   NH3
Calculate the mass of sodium amide needed to obtain 550 g of sodium azide, assuming there is a 95.0% yield of sodium azide.
Give your answer to 3 significant figures.
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(5)
(b)     If a car is involved in a serious collision, the sodium azide decomposes to form sodium and nitrogen as shown in the equation.
2NaN3(s)      2Na(s)   +   3N2(g)
The nitrogen produced then inflates the airbag to a volume of 7.50 × 10−2 m3 at a pressure of 150 kPa and temperature of 35 °C.
Calculate the minimum mass of sodium azide that must decompose.
(The gas constant R = 8.31 J K−1 mol−1)
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(6)
(c)     Sodium azide is toxic. It can be destroyed by reaction with an acidified solution of nitrous acid (HNO2) as shown in the equation.
2NaN3  +  2HNO2  +  2HCl    3N2  +  2NO  +  2NaCl  +  2H2O
(i)      A 500 cm3 volume of the nitrous acid solution was used to destroy completely 150 g of the sodium azide.
Calculate the concentration, in mol dm−3, of the nitrous acid used.
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(3)
(ii)     Nitrous acid decomposes on heating.
Balance the following equation for this reaction.
........HNO2     .......HNO3   +   .......NO   +   .......H2O
(1)
(d)     Sodium azide has a high melting point.
Predict the type of bonding in a crystal of sodium azide.
Suggest why its melting point is high.
Type of bonding .............................................................................................
Reason for high melting point ........................................................................
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(3)
(e)     The azide ion has the formula N3−
(i)      The azide ion can be represented as N  N − N−
One of these bonds is a co−ordinate bond.
On the following diagram, draw an arrowhead on one of the bonds to represent the direction of donation of the lone pair in the co−ordinate bond.
N  N − N−
(1)
(ii)     Give the formula of a molecule that has the same number of electrons as the azide ion.
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(1)
(iii)    Which is the correct formula of magnesium azide?
Tick (✓) one box.
 
 	Mg3N		
 	MgN		
 	MgN6		
 	Mg3N2		

(1)
(Total 21 marks)

Q19.(a)     Calcium phosphate reacts with aqueous nitric acid to produce phosphoric acid and calcium nitrate as shown in the equation.
Ca3(PO4)2    +    6HNO3    2H3PO4    +    3Ca(NO3)2
(i)      A 7.26 g sample of calcium phosphate reacted completely when added to an excess of aqueous nitric acid to form 38.0 cm3 of solution.
Calculate the concentration, in mol dm–3, of phosphoric acid in this solution.
Give your answer to 3 significant figures.
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(5)

(ii)     Calculate the percentage atom economy for the formation of calcium nitrate in this reaction.
Give your answer to 1 decimal place.
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(2)
(b)     Write an equation to show the reaction between calcium hydroxide and phosphoric acid to produce calcium phosphate and water.
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(1)
(c)     Calcium dihydrogenphosphate can be represented by the formula Ca(H2PO4)x where x is an integer.
A 9.76 g sample of calcium dihydrogenphosphate contains 0.17 g of hydrogen, 2.59 g of phosphorus and 5.33 g of oxygen.
Calculate the empirical formula and hence the value of x.
Show your working.
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(4)
(Total 12 marks)
Q20.Compounds containing Cu2+, OH– and CO32− ions are sometimes described as basic copper carbonates.
(a)     Solid Cu2(OH)2CO3 is added to an excess of dilute hydrochloric acid.
A solution of copper(II) chloride is formed, together with two other products.
(i)      Write an equation for the reaction.
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(2)

(ii)     Suggest one observation that could be made during the reaction.
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(1)
(b)     A 5.000 g sample of a different basic copper carbonate contains 0.348 g of carbon, 0.029 g of hydrogen and 1.858 g of oxygen.
(i)      State what is meant by the term empirical formula.
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(1)
(ii)     Calculate the empirical formula of this basic copper carbonate.
Show your working.
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(3)
(Total 7 marks)

Q21.Phosphoric(V) acid (H3PO4) is an important chemical. It can be made by two methods.
The first method is a two-step process.
(a)     In the first step of the first method, phosphorus is burned in air at 500 ºC to produce gaseous phosphorus(V) oxide.
P4(s)   +   5O2(g)   →   P4O10(g)
220 g of phosphorus were reacted with an excess of air.
Calculate the volume, in m3, of gaseous phosphorus(V) oxide produced at a pressure of 101 kPa and a temperature of 500 ºC.
The gas constant R = 8.31 J K–1 mol–1
Give your answer to 3 significant figures.
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(4)
(b)     In the second step of the first method, phosphorus(V) oxide reacts with water to form phosphoric(V) acid.
P4O10(s)   +   6H2O(l)   →   4H3PO4(aq)
Calculate the mass of phosphorus(V) oxide required to produce 3.00 m3 of 5.00 mol dm–3 phosphoric(V) acid solution.
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(3)

(c)     In the second method to produce phosphoric(V) acid, 3.50 kg of Ca3(PO4)2 are added to an excess of aqueous sulfuric acid.
Ca3(PO4)2(s)   +   3H2SO4(aq)   →   2H3PO4(aq)   +   3CaSO4(s)
1.09 kg of phosphoric(V) acid are produced.
Calculate the percentage yield of phosphoric(V) acid.
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(4)

(d)     Explain whether the first method or the second method of production of phosphoric acid has the higher atom economy.
You are not required to do a calculation.
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(1)
(Total 12 marks)

 

M1.D
[1]

M2.A
[1]

M3.C
[1]

M4.B
[1]

M5.D
[1]

M6.D
[1]

M7.C
[1]

M8.C
[1]

M9.B
[1]

M10.D
[1]

M11.          (a)     (i)      0.013;
1

(ii)     0.0065;
Answer to (i) ÷ 2
1

(iii)      = 84.3;
Allow 0.548 ÷ answer to (ii)
Allow 84.1 – 84.4
1

(b)     84.3 – 60 = 24.3;
1 mark for -60
1
          Mg;
          If 147.6 used the answer is 87.6 (1)
And this is Sr (1)
Allow consequential metal from their calculated A
Answer has got to be a metal to score M2
1
[5]

 

M12.          (a)     (i)      0.00301/ 3.01 × 10–3;
         Penalise < 3sf in (a)(i);
Allow 3.01 × 10–3 – 3.05 × 10–3.
(for candidates who have used Mg as 24)
1

(ii)     0.00602
Allow correct answer a(i) × 2.
1

(iii)     0.00965/ 9.65 × 10–3;
Allow 0.009646/ 0.0096-0.0097.
1

(iv)    0.00363 moles;
Allow range 0.0035 to 0.0037.
Allow (a)(iii) – 2 (a)(ii) (must be positive).
1

(b)     PV = nRT;
Allow all capitals/ lower case.
1
          V = ;
M2 Mark is for all numbers correct.
If units in answer are in dm3 allow this expression with 96 in denominator.
1
          0.0132 m3/ 13.2 dm3;
M3 Must have correct units/
allow 13200 cm3.
Allow min 2 sig figs in answer.
1

(c)     O = 69.6 (%);
1
                  2.17 : 4.35
Use of 7/8 CE then M1 only.
1
          (1 : 2) NO2
Mark for formula not ratio.
If NO2 and no working shown then allow 1 mark.
If 69.6% + NO2 only = 2.
Need to see evidence of M2 working.
Allow M2 conseq on the wrong M1 (ie max 1).
1
[10]

 

M13.          (a)     L =  (1) or 
must show working
          = 6.0225 × 1023 (1)
Ignore wrong units
NB answer only scores 1
2

(b)     equal (1)
Or same or 1:1
1

(c)     PV = nRT (or n = ) (1)
          =  (1)
          = 1.39 (1)
Allow 1.390 to 1.395
ignore units even if incorrect
answer = 1.4 loses last mark
3

(d)     0.732 ×  = 2.93 (1) mol.dm–3 (1)
OR M, mol/dm3, mol.l–1
allow 2.928 to 2.93
Note unit mark tied to current answer but allow unit
mark if answer = 2.9 or 3
2

(e)     (i)      moles H2SO4 =  × 1.24 = 0.0310 (1)
If use m1v1 = m2v2 scores 3 if answer is correct otherwise zero
         moles NH3 in 30.8 cm3 = 0.0310 × 2 = 0.0620  (1)
Mark is for ×2
CE if × 2 not used

         moles of NH3 in 1 dm3 = 0.620 ×  = 2.01 (1) (mol dm–3)
Allow 2.010 to 2.015
No units OK, wrong units lose last mark

(ii)     moles (NH4)2SO4 = moles H2SO4 = 0.310 (1)
Allow consequential wrong moles in part (i) if clear H2SO4=(NH4)SO4
Wrong formula for (NH4)2SO4 CE=0
         Mr (NH4)2SO4 = 132.1 (1)
Allow (132)
         mass = moles × Mr = 0.0310 × 132.1 = 4.10 (1)
if moles of (NH4)2SO4 not clear CE
(g) wrong unit loses mark
Allow 4.09 – 4.1 – 4.11
6

(f)      Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3
Formulae (1)
Balanced equation (1)
2
[16]

 

M14.          (a)     (i)      Mr = 132.1
1
132
0.0238
Allow 0.024
Allow 0.0237
Penalise less than 2 sig fig once in (a)
1

(ii)     0.0476
1
0.0474-0.0476
Allow (a) (i) × 2

(iii)     1.21
Allow consequential from (a) (ii)
ie allow (a) (ii) × 1000/39.30
Ignore units even if wrong
1

(b)     
Allow mass or Mr of desired product times one hundred divided by total mass or Mr of reactants/products
If 34/212.1 seen correctly award M1
1
= 16.0(3)%
Allow 16%
16 scores 2 marks
1

(c)     100(%)
Ignore all working
1

(d)     PV = nRT or n = 
If rearranged incorrectly lose M1 and M3
1
n = 
M2 for mark for converting P and T into correct units in any expression
1
= 0.59(4)
Allow 0.593
M3 consequential on transcription error only not on incorrect P and T
1

(e)     (Na2SO4)          H2O
(44.1%)           55.9%
M1 is for 55.9
1
44.1/142.1           55.9/18
0.310                   3.11
= 1                       = 10
Alternative method gives180 for water part = 2 marks
1
x = 10
X = 10 = 3 marks
10.02 = 2 marks
1
[13]

 

M15.(a)      34.0
Penalise precision once
1

(b)     1.76 mol dm–3
1

(c)     answer to (b) divided by 0.05
35(.3) on correct figures
1

Shows working
Correct answer only scores this mark
Lose this mark if any units are given for the factor
1
[4]

M16.(a)     pV = nRT
Do not penalise incorrect use of capitals / lower case letters.
Accept correct rearrangement of equation.
1

(b)     2C4H10 + 5O2 → 4CH3COOH + 2H2O
Accept any correct combination of multiples, including fractions.
1

(c)     23.0 g ethanol produces 30.0 g ethanoic acid
1

15.1% (4.54 ×100 / 30)
Do not penalise precision.
15.1% scores 2 marks.
Accept consequential answer on wrong mass of ethanoic acid for second mark only.
1
[4]

M17.          (a)     (i)      Mr MgO = 40.3
If used 40 then penalise this mark but allow consequential M2 (0.0185)
1
0.741/40.3 = 0.0184
0.018 with no Mr shown = 0
Penalise if not 3 sig figs in this clip only
1

(ii)     0.0184 × 5/2 = 0.0460
Allow 0.0459 to 0.0463
Allow their (a)(i) × 5/2 ie allow process mark of × 5/2 but insist on a correct answer being written down
Ignore sig figs
1

(b)     pV = nRT
1
(V =)
If rearranged incorrectly then lose M1
If this expression correct then candidate has scored first mark
0.0111
1
Ignore units
11.1 (dm3)
3 marks for 11.1 (dm3)
However if 11.1 m3 or cm3 allow 2 (ie penalise wrong units in final answer)
Ignore sig figs- but must be 2 sig figs or greater
1

(c)     (i)      0.0152 × 2 = 0.0304
Allow 0.03
1

(ii)     0.938 mol dm–3
Allow range 0.92 – 0.94
Minimum 2 sig figs
Allow consequential marking from (c)(i)
Ignore units even if wrong
1
[8]

 

M18.(a)     M1   550 ×  = 579 g would be 100% mass
Allow alternative methods.
There are 4 process marks:
1

M2   So  = 8.91 moles NaN3
or
M1    = 8.46 moles NaN3 (this is 95%)
M2   So 100% would be 8.46 ×  = 8.91 moles NaN3
1: mass ÷ 65
2: mass or moles × 100 / 95 or × 1.05
3: moles NaN3 × 2
4: moles NaNH2 × 39
1

Then M3 Moles NaNH2 = 8.91 × 2 = ( 17.8(2) moles)
1
M4   mass NaNH2 = 17.8(2) × 39
1
M5   693 or 694 or 695 (g)
If 693, 694 or 695 seen to 3 sig figs award 5 marks
1

(b)     M1   308 K and 150 000 Pa
1
M2   n =  or 
1
M3   = 4.4(0) or 4.395   moles N2
Allow only this answer but allow to more than 3 sig figs
1

M4   Moles NaN3 = 4.395 ×  (= 2.93)
M4 is for M3 × 
1
M5   Mass NaN3 = (2.93) × 65
M5 is for moles M4 × 65
1
M6 = 191 g
Allow 190 to 191 g allow answers to 2 sig figs or more
1

(c)     (i)      150 / 65 = 2.31 moles NaN3 or 2.31 moles nitrous acid
1
Conc = 2.31 × 
 
M2 is for M1 × 1000 / 500
1
4.6(1) or 4.6(2) (mol dm−3)
Only this answer
1
(ii)     3HNO2  HNO3 + 2NO + H2O
Can allow multiples
1

(d)     Ionic
If not ionic then CE = 0 / 3
1
Oppositely charged ions / Na+ and N3 − ions
Penalise incorrect ions here but can allow M3
1
Strong attraction between (oppositely charged) ions / lots of energy needed to overcome (strong) attractions (between ions)
M3 dependent on M2
1

(e)     (i)      N ≡ N  N−
Only
1
(ii)     CO2 / N2O / BeF2 / HN3
Allow other correct molecules
1
(iii)    MgN6
Only
1
[21]

M19.(a)     (i)      M1 - Mr calcium phosphate = 310(.3)
If Mr wrong, lose M1 and M5.
1

M2 - Moles calcium phosphate =        (= 0.0234)
 
0.0234 moles can score M1 and M2.
If Mr incorrect, can score M2 for .
 
Allow M2 and / or M3 to 2 significant figures here but will lose M5 if answer not 1.23.
1

M3 - Moles phosphoric acid = 2 × 0.0234 = 0.0468
Allow student’s M2 × 2. If not multiplied by 2 then lose M3 and M5.
1

M4 - Vol phosphoric acid = 0.038(0) dm3
If not 0.038(0) dm3 then lose M4 and M5.
1

Conc phosphoric acid = 
M5 = 1.23 (mol dm−3)
This answer only – unless arithmetic or transcription error that has been penalised by 1 mark.
Allow no units but incorrect units loses M5.
1

(ii)      × 100    OR     × 100
 
1 mark for both Mr correctly placed.
= 71.5%
2

(b)     3Ca(OH)2 + 2H3PO4  Ca3(PO4)2 + 6H2O
Allow multiples.
1

(c)     

If x = 2 with no working, allow M4 only.
Ca = 1.67 g (M1).
1

Mark for dividing by correct Ar in Ca and P (M2).
If M1 incorrect can only score M2.
1

Correct ratio (M3).
1

CaH4P2O8    OR    Ca(H2PO4)2    OR    x = 2
Value of x or correct formula (M4).
1

Alternative
Ca                   H2PO4
Ca = 1.67 g (M1).

Mark for dividing by correct Ar / Mr in Ca and H2PO4 (M2).
If M1 incorrect can only score M2.

Correct ratio (M3).
CaH4P2O8    OR    Ca(H2PO4)2    OR    x = 2
Value of x or correct formula (M4).
[12]

M20.(a)     (i)     H2O + CO2 (as products in any equation)
Allow H2O + H2CO3
1
Cu2(OH)2CO3 + 4HCl → 2CuCl2 + 3H2O + CO2
Allow multiples
Ignore states
1

(ii)     Bubbles or fizzing or effervescence
Or solid disappears
Or blue(-green) solution
Do not allow dissolves
Ignore CO2 gas or gas evolved
1

(b)     (i)      Simplest (whole-number) ratio of atoms of each element in a compound
Allow atoms of Cu, H & O in this compound
1
(ii)     Mass of copper = 2.765
Dividing masses by Ar
1
 
1
Correct whole number ratio of integers
or
Cu:C:H:O
3:2:2:8
or
Correct empirical formula Cu3C2H2O8
Any order
Ignore Cu3(OH)2(CO3)2
1
[7]

M21.(a)     Correct conversion of temperature and pressure (773 and 101 × 103)
Correct answer with or without working scores 4 marks
1
No moles P = (220 / 4 × 31.0) = 1.77
Max 2 (M1 and M3) if 31.0 used
(=0.451 m3 or if 220/31 rounded to 2 sf ie 7.1 then 0.452)
1
V = nRT/P (correct rearrangement or insert of values V = 1.77 × 8.31 × 773/101 × 103 =0.1128 m3)
Max 2 (M1 and M3) if 284 (P4O10) used then 0.0493
1
V = 0.113 (m3)
Must be 3 sig figs
1

(b)     No moles H3PO4 = 3 × 103 (dm3) × 5 = 15,000 (mols)
Correct answer with or without working scores 3 marks
If M1 incorrect then can only score M2
1
No moles phosphorus(V) oxide =  ( = 3,750 mols)

If M2 incorrect can only score M1
1
1.1 × 106 or1.07 × 106 or 1.065 × 106 (g)
or 1,100 or 1,070 or 1065 kg
or 1.1 or 1.07 or 1.065 tonne
= (3.75 × 103 × 284.0)
Min 2 sig fig
1

(c)     No moles Ca3(PO4)2 (= 3.50kg =)  = 11.28
Correct answer with or without working scores 4 marks
If M1 incorrect can only score M2 and M3
1
Theoretical No. moles H3PO4 = 11.28 × 2 = 22.56
If M2 incorrect can only score M1 and M3
1
Theoretical mass H3PO4 = 22.56 × 98(.0) = 2211
If M3 incorrect can only score M1and M2
1
or Actual No. moles H3PO4 produced =  = 11.12
49 – 49(.312) (%)

1

(d)     Method 1 / (a) & (b) because only one product / no other products formed / atom economy = 100% (even though two steps)
Allow calculations
Do not allow if P2O5 is formed
Allow converse explanation
1
[12]