chapter 8 salt part 4

CHAPTER 8SALT

NURUL ASHIKIN BT. ABD RAHMAN PART 4

“Manusia mampu mengemukakan 1000 alasan mengapa mereka gagal tetapi mereka sebenarnya hanya perlukan satu sebab yang

kukuh untuk berjaya...’’

NaCl

BaSO4

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

Use Precipitation Method

Acid + Alkali salt + water

Titration method

Evaporation/Heating

Cooling/crystallization

Filtration

Dry

LEARNING OUTCOMES

Solve problems involving calculation of quantities of reactants or products in stoichiometric reactions.

STOICHIOMETRIC REACTION(Balance Chemical Equation)

A balanced chemical equation provide information about the

number of moles of each reactant and the product in the reaction.

Mass (g)

No. of moles, n

Volume of solution (dm3)

¸Molarmass

molarityX molar mass X molarity

Calculation Step

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]

EXAMPLE 1:

Ans: 17.04 g

Step 1:Al2O3 + 6HNO3 2Al(NO3)3 + 3H2O

Solution:

Step 2:Mass Al2O3 = 4.05 g

Mass Al(NO3)3 = ?

Step 3:From chemical equation1 mol Al2O3 2 mol Al(NO3)3

Step 4:

molar mass Al2O3 = 27(2) + 16(3) = 102 g mol-

Mole Al2O3 = 0.04 mol

2 3

mass (g)mole Al O =

molar mass ( )g mol

2 3

4.05

102

gmole Al O

g mol

Step 5:0.04 mol Al2O3 produced

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

3 3

0.04 2Al(NO )

1

3 30.08 mol Al(NO )

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]

EXAMPLE 2:

Ans: 100 cm3

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.

Solution

Step 4: 10 g of CaCO3

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

3

3 3 3

. 0.2

2

0.1 0.1 1000 100

no of mole HCl mol

molarity of HCl mol dm

dm cm cm

10 10

40 12 3(16) 100

0.1 mol

GROUPDISCUSSION

POP QUIZ

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

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.

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.

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

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

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.

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.

End of slide… Thank you..