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