chapter 8 salt part 4
DESCRIPTION
Chemistry Form 4: UPSI/SLISS 2012 Chapter 8: Salt (Part 4)TRANSCRIPT
![Page 1: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/1.jpg)
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...’’
![Page 2: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/2.jpg)
NaCl
BaSO4
![Page 3: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/3.jpg)
• Choose soluble salt solution containing anion and cation insoluble salt.• Mix the two solution.• Filter.• Wash.• Dry the precipitate.
Use Precipitation Method
![Page 4: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/4.jpg)
Acid + Alkali salt + water
Titration method
Evaporation/Heating
Cooling/crystallization
Filtration
Dry
![Page 5: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/5.jpg)
LEARNING OUTCOMES
Solve problems involving calculation of quantities of reactants or products in stoichiometric reactions.
![Page 6: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/6.jpg)
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.
![Page 7: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/7.jpg)
Mass (g)
No. of moles, n
Volume of solution (dm3)
¸Molarmass
molarityX molar mass X molarity
![Page 8: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/8.jpg)
Calculation Step
![Page 9: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/9.jpg)
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
![Page 10: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/10.jpg)
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
![Page 11: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/11.jpg)
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
![Page 12: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/12.jpg)
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 )
![Page 13: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/13.jpg)
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
![Page 14: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/14.jpg)
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
![Page 15: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/15.jpg)
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
![Page 16: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/16.jpg)
GROUPDISCUSSION
![Page 17: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/17.jpg)
POP QUIZ
![Page 18: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/18.jpg)
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
![Page 19: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/19.jpg)
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.
![Page 20: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/20.jpg)
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.
![Page 21: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/21.jpg)
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
![Page 22: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/22.jpg)
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
![Page 23: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/23.jpg)
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.
![Page 24: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/24.jpg)
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.
![Page 25: Chapter 8 salt part 4](https://reader034.vdocuments.net/reader034/viewer/2022051312/546a677faf79596c268b456e/html5/thumbnails/25.jpg)
End of slide… Thank you..