Author: nashton
Company: RM plc

F321 module 1 Practice 3:

1.	M(BaO) = 137 + 16 = 153 
moles BaO = 500/153 or 3.268 mol 
moles Ba = 3.268/2 or 1.634 
mass Ba formed = 1.634 × 137 = 224 g 
accept 223.856209/223.86/223.9 g.
if 6 mol BaO forms 3 mol Ba, award 3rd mark
Alternative method
mass 6BaO=918 g 
mass 3Ba = 411 g 
1g BaO forms 411/918 g Ba 
500 g BaO forms 223.856209/223.86/223.9 g Ba 
[4]

2.	(i)	

		protons	neutrons	electrons
	^25Mg	12	13	12	
^26Mg	12	14	12		2
(ii)	1s^22s^22p^63s^2
24 × 78.60/100 + 25 × 10.11/100 + 26 × 11.29/100	1
(iii)	= 24.33  (calc value: 24.3269. This scores one mark)
24.32 with no working, award 1 mark only.
24.3 with no working, no marks (Periodic Table value)	2
[5]

3.	(i)	Oxidation state goes from 0 in O_2 
→ −2 in MgO 	2
(ii)	

or with Mg full shell.
correct dot and cross; correct charges	2
[4]

4.	(i)	MgO has reacted with CO_2	1
(ii)	Solid dissolves / disappears
Fizzing / bubbles	2
MgO + 2HCl → MgCl_2 + H_2O
MgCO_3 + 2HCl → MgCl_2 + CO_2 + H_2O
both reactions form magnesium chloride/MgCl_2	3
[6]

5.	(i)	ratio N : H : S : O = : : : :
= 2 : 8 : 1 : 3
Empirical formula = N_2H_8SO_3
	N_2H_4SO_3 is worth 1 mark from consistent use of at nos.	2
(ii)	H_2O + 2NH_3 + SO_2 → (NH_4)_2SO_3	1
(Award mark for N_2H_8SO_3)
[3]

6.	(a)	(i)	Amount of substance that has the same number of
	particles as there are atoms in 12 g of ^12C/
	6 × 10^23/ Avogadro’s Number	1
(ii)	moles =  = 0.0330 mol
	moles Cl_2 =  = 0.0165 mol	1
(iii)	volume Cl_2 = 0.0165 × 24000 = 396 cm^3 / 0.396 dm^3
	792 cm^3 worth 1 mark (no molar ratio)
	1584 cm^3 worth 1 mark (x 2)
	units needed.	2
(iv)	bleach / disinfectant /sterilising /killing germs	1
 
(b)	NaClO_3	1
[6]

7.	(i)	CaCO_3 (s) + 2HCl(aq) → CaCl_2(aq) + CO_2(g) +H_2O (l)
CaO(s) + 2HCl(aq) → CaCl_2(aq) + H_2O (l)
each balanced equation 1 mark (2)
all state symbols (1)	3
(ii)	CaCO_3 fizzes/ gas given off/ gas evolved / carbon dioxide evolved (1)	1
[4]

8.	(i)	(atoms of) same element/same atomic number/number of protons with
different numbers of neutrons/diff masses	1
(ii)	
	proton	neutron	electron
relative mass 1	1	/	negligible
relative charge	+1	0	−1	
i.e. 1 mark for each correct row
for electron, accept 1/1500 – 1/2000
for charges, accept +; 0; –
		2
[3]

9.	(i)	average atomic mass/weighted mean/average mass 
compared with carbon-12 
1/12th of mass of carbon-12/on a scale where carbon-12 is 12 
OR
(ii)	The mass of 1 mole of atoms of an element
compared with 12 g  of carbon-12 	3
A_r =  / 121.8558 
= 121.9 	2
[5]

10.	(i)	Mass Sb_2S_3 in stibnite = 5% of 500 kg = 25.0 kg 
Moles Sb_2S_3 =  / 73.5/ 73.529 /73.53/ 74 mol 
		(calculator value: 73.52941176)
If 5% is not used, 1471 mol; ecf for 2nd mark
		(calculator value: 1470.588235)
If 5% is used 2nd, 73.6 mol: OK for both marks	2
(ii)	moles Sb = 2 × 73.5 mol 
mass Sb = 2 × 73.5 × 122 g = 17.9 kg 
If the 2 isn’t used, answer = 73.5 × 122 = 8.95 
ecf ans from (i) x 2
ecf ans above x 2
OR
% Sb = 244/340 = 71.7% 
mass Sb = 25.0 × 71.7/100 = 17.9 kg  (ecf as above)	2
[4]

The King's CE School