Chemistry Form 4: UPSI/SLISS 2012 Chapter 8: Salt (Part 4)

1. “Manusia mampu mengemukakan 1000
alasan mengapa mereka gagal tetapi mereka
sebenarnya hanya perlukan satu sebab yang
kukuh untuk berjaya...’’
NURUL ASHIKIN BT. ABD RAHMAN PART 4

2. NaCl
BaSO4

3. Use Precipitation Method
• Choose soluble salt solution
containing anion and cation
insoluble salt.
• Mix the two solution.
• Filter.
• Wash.
• Dry the precipitate.

4. Acid + Alkali  salt + water
Titration method
Evaporation/Heating
Cooling/crystallization
Filtration
Dry

5. LEARNING OUTCOMES
 Solve problems involving calculation
of quantities of reactants or
products in stoichiometric reactions.

6. A balanced chemical equation
provide information about the
number of moles of each reactant
and the product in the reaction.

7. Volume of
Mass (g)
solution (dm3)
molarity
Molar X molar
mass mass X molarity
No. of moles, n

8. Calculation Step
Write balanced equation
Write information from question
Write information from chemical equation
Change information from question into mole
Use the relationship between number of mole from
chemical equation
Change the information to the unit required

9. EXAMPLE 1:
4.05 g of aluminium oxide powder is mixed
with excess dilute nitric acid and the mixture
is heated. Calculate the mass of aluminium
nitrate produced.
[RAM: N,14; O,16; Al, 27]
Ans: 17.04 g

10. Solution:
Step 1:
Al2O3 + 6HNO3  2Al(NO3)3 + 3H2O
Step 2:
Mass Al2O3 = 4.05 g
Mass Al(NO3)3 = ?
Step 3:
From chemical equation
1 mol Al2O3 2 mol Al(NO3)3

11. Step 4:
mass (g)
mole Al2 O3 =
molar mass ( g mol )
molar mass Al2O3 = 27(2) + 16(3) = 102 g mol-
4.05 g
mole Al2O3
102 g mol
Mole Al2O3 = 0.04 mol

12. Step 5:
0.04 2
0.04 mol Al2O3 produced Al(NO3 )3
1
0.08 mol Al(NO3 )3
Step 6:
Mass Al(NO3)3 = ?
Molar mass Al(NO3)3 = 27 + [14+16(3)]3
= 213 g mol-
Mass Al(NO3)3 = mol x molar mass
= 0.08 mol x 213 g mol-
= 17.04 g

13. EXAMPLE 2:
What is the volume of 2.0 mol dm-3
hydrochloric acid required to dissolve 10 g of
marble ( calcium carbonate)?
[RAM: H,1 ; O,16; C,12; Ca,40]
Ans: 100 cm3

14. Solution
Step 1: CaCO3 + 2HCl  CaCl2 + CO2 + H2O
Step 2: Molarity HCl = 2 mol dm-3
Mass CaCO3 = 10 g
Volume HCl = ?
Step 3: from chemical equation, 1 mol CaCO3 react
with 2 mol HCl to complete reaction.

15. 10 10
Step 4: 10 g of CaCO3
40 12 3(16) 100
0.1 mol
Step 5: hence 0.1 mol CaCO3 requires 0.1 x 2 = 0.2
mol HCl for a complete reaction.
Step 6: Volume HCl = molarity x volume
no. of mole HCl 0.2 mol
3
molarity of HCl 2 mol dm
0.1 dm3 0.1 1000 cm3 100 cm3

18. Question 1
50 cm of 2 mol dm–3 sulphuric acid is added to an
excess of copper(II) oxide powder. Calculate the
mass of copper(II) sulphate formed in the
reaction. [Relative atomic mass: H , 1; O ,16;
Cu,64; S,32].
Ans:16 g

19. Question 2
A student prepared some copper (II) nitrate by
reacting copper (II) oxide with excess nitric
acid. How many grams of copper (II) nitrate
will be produced, if 40 g of copper (II) oxide is
used in the reaction? [Cu,64; N, 14; O,16].
Ans:94 g Cu(NO3)2.

20. Question 3
27.66 g of lead(II) iodide is precipitated when
2.0 mol dm–3 of aqueous lead(II) nitrate
solution is added to an excess of aqueous
potassium iodide solution. Calculate the
volume of aqueous lead (II) nitrate solution
used. [Relative atomic mass: I, 127; Pb,207].
Ans: 30cm3.

21. Question 4
Calculate the number of moles of aluminium
sulphate produced by the reaction of 0.5 mol
of sulphuric acid with excess aluminium
oxide?
Ans: 0.167 mol

22. Question 5
150 cm3 of 1.0 mol dm-3 ammonia solution is
completely neutralised with phosphoric acid
using a titration methode. Calculate the mass
of ammonium phosphate formed. [RAM: H,1 ;
N,14; O,16; P,31] .
Ans: 7.45 g

23. Question 6
What is the mass of zinc oxide when Zinc
oxide powder is added to 100 cm3 of 2 mol
dm-3 nitric acid to form zinc nitrate. Then
calculate the mass of zinc nitrate produced.
[Relative atomic mass: H,1; O, 16; Cl,35.5,
Zn,65; N, 14].
Ans: 8.1g ZnO; 18.9 g Zn(NO3)2.

24. Question 7
Copper (II) sulphate is prepared by added 5.6
g of copper (II) oxide to 1.25 mol dm-3
sulphuric acid. Calculate the volume of acid
needed to react completely with the copper
(II) oxide. [ Relative atomic mass: O,16; Cu,64].
Ans: 56 cm3.