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CONCENTRATION UNITS Mass of solute + Mass of solvent =Mass of solution 20g NaOH dissolved in H 2 O to form 120g solution. Mass of solute = 20g Mass of solvent = 120-20 =100g

Concentration ( g/L ) = Mass solute(g) / V solution ( L ) 90g glucose is dissolved to form 500ml solution . = 90 g / 0.5 L 180 g/L

EXERCISE: Give the concentration of a AgNO 3 solution with 5.00 g of solute dissolved in 250.0 mL solution. [Ans: 20.0 gL -1 ]

Molarity ( Mol/L @ M ) = Mole solute / V solution ( L ) 90g glucose,C 6 H 12 O 6 dissolved to form a 500ml solution ( 90 g/180 g mol -1 ) / 0.5 L= 1.0M

Exercise: What is the molarity of 85.0 mL of ethanol, C 2 H 5 OH that Contains 1.77 g of ethanol? Ans: 0.453 M 2. Find the molarity of 9.80g of hydrogen sulphate,H 2 SO 4 dissolved in 0.250 litre of solution. Ans: 0.400 M

% w/v ( weight per volume ) Commonly used in drugs eg 20%w/v antihistamine; Thioridazine HCl = Mass solute(g) / V solution@sample (ml) @ kg / L or equivalent unit. X 100 Linctus Codeine Paed.BP(British Pharmacopeia) contains 15mg codeine in 0.005L solution.Calculate concentration in %w/v . * [0.3 % w/v ]

Exercise: 1. Dextromethorphan, a cough mixture, contains 3.0 % w/v of Dextromethorphan hydrobromide with 0.2 % of methylparaben as one of the preservatives What is the mass in gram of Dextromethorphan hydrobromide in 90 mL of the cough mixture? ANS: 2.7 g

Molality ( mol/kg ) = Mole solute / mass solvent ( kg ) 80g NaOH dissolved in water to form 200g solution. Na =23 Mass solvent = 200-80 = 120g Molality =(80/40) / (120/1000) = 16.67 mol/kg

EXERCISE: 1. A KOH solution containing 7.00 g KOH (RAM KOH = 56.0)dissolved in 500.0g water to form a 502.0 mL solution. Find its molality. Ans: 0.25 m

% by mass(w/w) = Mass solute / mass solution x 100 COMMONLY USED IN DRUGS eg 2% w/w hydroquinone ( steroid ) 80g NaOH dissolved in water to form 200g solution. Na =23 80 / 200 x 100 = 40% w/w

Exercise: Calculate the percent by mass of the solute in an aqueous solution containing 5.50 g of NaBr in 78.2 g of solution. Ans:7.03% Calculate the amount of water (in grams) added to 5.00 g of urea, (NH 2 ) 2 CO to prepare 16.2 percent by mass solution. Ans: 25.9 g

Parts per million (ppm) = (Mass solute/mass sample) x 10 6 Commonly used in ENVIRONMENTAL Applications Units for ppm = mg/L @  g/mL @ mg/kg There are 20mg Hg 2+ ions in a 100g river sample = (0.02 g/ 100 g) x 10 6 = 200 ppm

EXERCISE: 1. The concentration of calcium ions in blood is 100.0 ppm. Calculate the mass of calcium ions in 500.0 g of blood. Ans: 0.05 g

2. What is the volume of 1.55 ppm solution formed by dissolving 454 g of NaF in water? Ans: 2.93 x 10 5 L C ppm = mass of solute (mg ) volume of solution (L) 1.55 (mg/L) = 454 g x 10 3 mg/g volume of solution (L) volume of solution = 454 x 10 3 mg 1.55 mg/L = 2.93 x 10 5 L

Exercise: 1.Calculate the mass of NaCl used to make a salt 1.00 ppm solution of 500 mL ? Ans: 5.00 x 10 -4 g Ans: mass of solute (mg ) = 1.00 mg L -1 0.500 L mass of NaCl = 0.500 L x 1.00 mg L -1 = 0.500 mg = 0.500 mg x 10 -3 g/mg = 5.00 x 10 -4 g

Mole Fraction 80g NaOH dissolved in water to form 200g solution. Na =23 Mass solute ( NaOH ) = 80g Mass solvent ( water ) = 200-80 = 120g

Mole Fraction Mole NaOH = 80 g/40 g mol -1 = 2.0 mol Mole solvent / water = 120 g/18 gmol -1 = 6.667 mol Mole fraction NaOH = n NaOH (n NaOH + n H 2 O) = 2.0 mol/ 8.667 mol = 0.231

Mole fraction Mole fraction H 2 O = n H 2 O ( n H 2 O + n NaOH ) = 6.667/8.667 = 0.769 @ TOTAL mole fraction =1 Mole fraction H 2 O = 1-0.231 = 0.769

Exercise: What is the mole fraction of of each component in a solution containing 1.0 mole of SO 2 in 1.5 mole of water? Ans: 0.40; 0.60

Dilution C 1 V 1 = C 2 V 2 C 1 = Initial [ ] C 2 = Final [ ] V 1 = Initial volume V 2 = Final/total volume 500ml water is added to 2.0M 1.5L antibiotic , amoxycillin solution. Calculate the final molarity

Dilution Solution C 1 V 1 = C 2 V 2 2.0 x 1.5 = C 2 x 2.0 C 2 = 1.50M

Exercise: Calculate the volume of water added to 0.50 M ethanoic acid to make a 450 mL 0.10 M solution? Ans: 360 mL 2. What is the concentration of the original solution that has been diluted with 100 mL water to form a 500 mL diluted solution of 0.25 M? Ans: 0.31 M

1. A solution contains 55 g toluene, C 7 H 8 and 55 g of bromobenzene,C 6 H 5 Br. Find the mole fraction of each component. Ans: 0.63; 0.37 2. The concentration of Cl - ion in a water source is about 0.0 1 g per 1000g of water. Find the concentration of Cl - ion in the water in parts per million, ppm. Ans: 10.0 ppm 3. Express the concentration of a 0.10 M HCl solution in g/L .[H = 1.00; Cl = 35.5 ] [Ans:3.65 g/L]

4. A solution contains 3.85 g H 2 S in 100 g of water at 20 0 C. What is the percent mass of this solution? Ans: 3.71 % How many grams of NaOH and water are needed to prepare 250.0 g of of 1.00 % NaOH solution? Ans: 2.50 g, 2.47.5 g 6. Calculate the volume of water added to 20 g NaOH to make a 450 mL 0.20 M solution Ans: 270 mL

7 . How many moles of NaOH are required to prepare a solution of 250 mL with 0.010M ? How many grams of NaOH are required ? If the solvent used is 249 mL , find its molality

Oxidation number ( ON ) Rules ON of elements eg Na, Cl 2 = 0 ON of ions = charge of the ion eg. Fe 3+ ion ON = +3 ON of H & O in compounds = +1 & -2 eg. ON of H in H 2 O = +1 ON of Group 1,2,13 elements eg. Na,Mg & Al in compounds = +1,+2 &+3.

Oxidation Number ( ON ) ON of all atoms in molecules/ions = overall charge of the molecule *ON of H in metal hydrides = -1 eg Na H *ON of O in peroxides = -1 eg H 2 O 2 , BaO 2 Examples a) Mn O 4 - x + 4(-2) = -1 ON of Mn = x x = +7

Oxidation Number ( ON ) Examples b) N H 3 x +3(1) = 0 x = ON of N x = -3 c) Cl O 3 - x + 3(-2) = -1 x = ON of Cl x = +5 d) Pb (NH 3 ) 4 2+ x + 4(0) = +2 x = ON of Pb x = +2

State the charge and oxidation number of atom underlined in the the following substances: Na , Zn 2+ , N 2 , C O 2 , H 2 S , Ca H 2 2 Assign oxid. No. to the underlined elements: Mn O 4 - Mn O 2 Ans: +7 +4 S 2 O 3 2- SO 4 2- Ans: +2 +6

Balancing Chemical Equations Inspection method ( Just like SPM !! ) Exercises i) Mg + HCl  MgCl 2 + H 2 ii) C 4 H 10 + O 2  CO 2 + H 2 O iii) Al + O 2  Al 2 O 3 iv) Cu + AgNO 3  Cu(NO 3 ) 2 + Ag

Review of last lecture 1.Concentration ( g/L ) 2. Molarity ( Mol/L @ M ) 3.Molarity ( Mol/L @ M ) 4. % by mass(w/w) 5. % w/v ( weight per volume )

7. Mole Fraction 6.. Parts per million (ppm) 8. Dilution 9. Oxidation number ( ON ) Balancing Chemical Equations By inspection Ion-electron method

Lecture 7& 8 Objectives: Balance redox equations in acidic solutions by ion-electron method. Solve stoichiometric calculation problems using mole concepts and concentrations units. Perform stoichiometric calculations involving limiting and excess reactants. Solve calculation problems involving normal and redox titrations.

Balancing Redox Equations Redox rxn = [ O ] and [ R ] simultaneously Normally involves IONIC equations [ O ] = Oxidation number INCREASES [ R ] = Oxidation number DECREASES

Balancing Redox Equations Example 1 Fe 2+ + Cr 2 O 7 2-  Fe 3+ + Cr 3+ Step 1 Calculate ON of all atoms, identify [ O ] and [ R ]  Separate to 2 half equations Fe 2+ + Cr 2 O 7 2-  Fe 3+ + Cr 3+ ON +2 +6 -2 +3 +3

Balancing Redox Equations Step 1 [ O ] Fe 2+  Fe 3+ [ R ] Cr 2 O 7 2-  Cr 3+ Step 2 Balance atoms that change ON ie Fe & Cr [ O ] Fe 2+  Fe 3+ [ R ] Cr 2 O 7 2-  2 Cr 3+ Step 3 Balance O atoms by placing H 2 O on the side with less O & balance the O atoms. [ O ] Fe 2+  Fe 3+ [ R ] Cr 2 O 7 2-  2Cr 3+ + 7H 2 O

Balancing Redox Equations Step 4 Balance H by placing H + on the side with less H [ O ] Fe 2+  Fe 3+ [ R ] 14H + + Cr 2 O 7 2-  2Cr 3+ + 7H 2 O Step 5 Calculate charges of ions/molecules/atoms & place electrons on the side with HIGHER +ve charge [ O ] Fe 2+  Fe 3+ + e [ R ] 14H + +Cr 2 O 7 2- + 6e  2Cr 3+ + 7H 2 O

Balancing redox equations Step 6 Balance the no. of electron & cancel appropriately [ O ] Fe 2+  Fe 3+ + e ( x6 ) [ R ] 14H + +Cr 2 O 7 2- + 6e  2Cr 3+ + 7H 2 O [ O ] 6Fe 2+  6Fe 3+ + 6e [ R ] 14H + +Cr 2 O 7 2- + 6e  2Cr 3+ + 7H 2 O ______________________________________ 6Fe 2+ + 14H + + Cr 2 O 7 2-  6Fe 3+ + 2Cr 3+ + 7H 2 O _________________________________

Exercises Cl 2 + C 2 O 4 2-  Cl - + CO 2 b) SO 3 2- + MnO 2 + H +  SO 4 2- + Mn 2+ Answers Cl 2 + C 2 O 4 2-  2Cl - + 2CO 2 b) SO 3 2- + MnO 2 + 2H +  SO 4 2- + Mn 2+ + H 2 O

SOLUTION: a. Cl 2 + C 2 O 4 2-  Cl - + CO 2 Step 1 : Cl 2 + C 2 O 4 2-  Cl - + CO 2 ON 0 +3 -2 -1 +4 -2 Step 1 +2 : Cl 2  2 Cl - [R] C 2 O 4 2-  2 CO 2 [O] Step 3 + 4 : Cl 2 -  2 Cl - C 2 O 4 2-  2 CO 2

Step 5 : Cl 2 +2e  2 Cl - --------- I) C 2 O 4 2-  2 CO 2 +2e--------- ii) Step 6: I) + ii) Cl 2 + C 2 O 4 2-  2Cl - + 2CO 2 b)SO 3 2- + MnO 2 + H +  SO 4 2- + Mn 2+ + H 2 O Steps: 1 S O 3 2- + Mn O 2  S O 4 2- + Mn 2+ +4 +4 +6 +2

Step 1&2 SO 3 2-  SO 4 2- [O] MnO 2  Mn 2+ [R] Step 3 SO 3 2- + H 2 O  SO 4 2- [O] MnO 2  Mn 2+ + 2 H 2 O [R] Step 4 SO 3 2- + H 2 O  SO 4 2- + 2H + [O] MnO 2 + 4H +  Mn 2+ +2 H 2 O [R] STEP 5 SO 3 2- + H 2 O  SO 4 2- + 2H + + 2e [O] ---I) MnO 2 + 4H + + 2e  Mn 2+ +2 H 2 O [R]– ii) Step 6 i) + ii) SO 3 2- + MnO 2 + 2H +  SO 4 2- + Mn 2+ + H 2 O

Stoichiometry Involves calculations of mass , mole, no. of particles, molar gas volume & concentration units such as molarity, g/L in EQUATIONS Example 1 2Mg + O 2  2MgO Calculate mass of MgO formed when 4.0 moles of oxygen is reacted

Stoichiometry Solution 1 mole O 2  2 mole MgO 0.4 mole O 2  0.4 x 2 = 0.8 mole MgO Mass MgO = 0.8 x (24+16) = 32.0g

Stoichiometry Example 2 2Mg + O 2  2MgO Calculate volume of O 2 at STP required to react with 12 x 10 22 Mg atoms . 1 mole O 2 requires 2 mole Mg Mole Mg = 12 x 10 22 / 6 x 10 23 =0.2 Mole O 2 = 0.2 x 2 = 0.4 Volume O 2 STP = 0.4 x 22.4 = 8.96L

Limiting & Excess Reactant Limiting reactant( LR ) = A reactant which is completely used up & determines the quantity of products. Excess reactant( ER ) = A reactant which is NOT completely used up &DOES NOT determine the quantity of products.

Limiting & Excess Reactant Example 1 2Mg + O 2  2MgO Calculate mass of MgO formed when 4.0 moles of oxygen is reacted. Limiting reactant is oxygen gas , O 2 because ONLY oxygen gas is given.

Exercise: For the following reactions: KClO 3 (p) KCl(p) + O 2 (g) Find the mole of oxygen formed if 0.25 mole of KClO 3 is heated to produce it. What is the mass of O 2 ? b) NO 2 (g) + H 2 (g) HNO 3 (l) + NO (g) Calculate the mass of hydrogen used to react with NO 2 to produce 1.5 mole of HNO 3 .

2KClO 3 (s) 2KCl(s) + 3O 2 (g) 2 mol 3 mol mole of O 2 =(3/2) x 0.25 = 0.38 mole mass of O 2 = 0.38 mole x 32.0 g mole -1 4NO 2 (g)+ H 2 (g) 2HNO 3 (l)+ 2NO (g) 1 mol 2 mol mole of H 2 = 1.50 /2 = 0.75 mole mass of H 2 = 0.75 mol x 2.00 g mol -1 = 1.50 g

Limiting & Excess Reactant Example 2 Mg + 2HCl  MgCl 2 + H 2 If 48g Mg is reacted with 6 x 10 23 molecules of HCl, determine LR and ER . Mg = 24 Thus, calculate mass of hydrogen gas formed. Determine mass of ER left at the end of the rxn .

Limiting & Excess Reactant Solution 1 st step = Determine LR and ER Mole Mg = 48/24 = 2.0 Mole HCl = 6 x 10 23 / 6 x 10 23 = 1.0 Mg HCl 2.0/1 =2.0 >> 1.0/2 = 0.5 << HCl is the LR …mole/coef ratio is SMALLER Mg is the ER…. mole/coef ratio is LARGER

Limiting & Excess Reactant Solution Since HCl is the LR, it is used to calculate mass of hydrogen ( PRODUCT ) n HCl = 1.0 mol n H 2 = 1.0 / 2 = 0.5 mol Mass of H 2 = 0.5 mol x 2 g mol -1 = 1.0g

Limiting & Excess Reactant Solution n of ER( Mg ) left =n Mg supplied – n Mg used (that reacts with LR) = 2.0 – (1.0 / 2 ) = 1.5 mole Mass of ER (Mg) left = 1.5 x 24 = 36.0g

Exercise your brain muscle: Zn(s) + S(s) ZnS(s) 6.35 g Zn was reacted with 8.0 g sulphur to form a white power, ZnS.[ Zn = 65.4; S = 32.0 ] Calculate the mole of each reactant Determine the limiting reactant and the excess reactant

iii) Find the mole of ZnS formed and hence deterrmine its mass. iv) Find the mole of excess reactants left over after the reaction is complete. v) From iv) determine the mass of excess reactant left over.

a) Mole of Zn = 6.35 g / 65.4 g mol -1 =0.0971mol mole of S = 8.0 g / 32.0 g mol -1 = 0.25 mol b) To find limiting reagent: Zn S ratio of n / the stoi.coefficient : 0.0971/1 : 0.25/1 = 0.0971 : = 0.25 The limiting reactant is Zn with the smaller ratio The excess reactant is S Solution

c) From equation, 1 mol Zn ≡ 1 mol of ZnS 0.0971 mol Zn ≡ 0.0971 mol ZnS = 0.0971 mol x [65.4 + 32.0] g mol-1 = 9.46 g d) 1 mol of Zn (limiting react ant) ≡ 1mol S (excess reactant) 0.0971 mol Zn ≡ 0.0971 mol S e) Excess S left over after the reaction = 0.25 – 0.0971 = 0.15 3mol S mass of excess S left over = 0.153 mol x 32.0 g mol-1 = 4.89 g

% Yield % yield = actual yield / theoretical yield x 100 2Mg + O 2  2MgO Calculate mass of MgO formed when 4.0 moles of oxygen is reacted. Mg = 24 Actual yield = Practically obtained from experiment !!! Theoretical yield = MAX yield of MgO expected

% yield Solution n O 2 = 4.0 n MgO = 4.0 x 2 = 8.0 Theoretical yield of MgO = 8.0 x 24 = 192.0g *Given 150g MgO obtained in the EXPT % yield = 150 /192 x 100 = 78.125%

Exercise CaCO 3 + 2HCl  CaCl 2 + H 2 O + CO 2 100g CaCO 3 reacts with 2.0M 400ml HCl, determine volume of CO 2 released at STP . Calculate mass of ER left at the end of the reaction. Calculate % yield if 7.20L CO 2 is obtained in the experiment. Answers : 8.96L, 60g, 80.36%

a) Mole of CaCO 3 = 100g /100.0 g mol -1 = 1.00 mol mole of HCl = 2.0 M x 0.400L = 0.80 mol b) To find limiting reactant: CaCO 3 HCl ratio of n / the stoi.coefficient : 1.00/1 : 0.80/2 = 1.00 : = 0.40 The limiting reactant is HCl with the smaller ratio The excess reactant is CaCO 3 Solution

d) 2 mol of HCl (limiting react ant) ≡ 1mol CO 2 0.80 mol HCl ≡ 0.80/2 mol CO 2 = 0.40 mol Volume of CO 2 released at STP = 0.40 mol x 22.4 L mol -1 =8.96 L e) Excess CaCO 3 left over after the reaction = 1.00 - 0.40 = 0.60 mol mass of excess CaCO 3 left over = 0.60 mol x 100.0 g mol-1 = 60 g

Titrations There are 2 types : Normal Titration - indicator is used eg. NaOH + HCl  NaCl + H 2 O Redox titration - indicator is not required since there is already a colour change eg. Cl 2 + C 2 O 4 2-  2Cl - + 2CO 2 (yellowish brown) (C’less)

Titrations A procedure to determine concentration of a solution using another solution of KNOWN concentration called STANDARD solution. Titre = Solution added from burette End point = The point where indicator changes colour Equivalence point = The point where mole of H + = mole of OH -

Normal Titration Example 1 2.0M 100ml KOH is required to neutralise 50ml H 2 SO 4 . Calculate molarity of H 2 SO 4 . 2KOH + H 2 SO 4  K 2 SO 4 + 2H 2 O Method 1 M a V a / M b V b = a/b M a = molarity of acid, M b = molarity of base V a = volume of acid, V b =volume of base

Normal Titration M a x 50 / 2.0 x100 = 1 / 2 M a = 2.0M @ mol/L Method 2 Mole KOH = 2 x100 / 1000 = 0.2 Mole H 2 SO 4 = 0.2 / 2 = 0.1 Molarity of H 2 SO 4 = 0.1 / 0.05 = 2.0M

Exercise 1) 196 g/L x ml sulphuric acid reacts with 1.5M 400 ml sodium hydroxide.Calculate value of x. * [ 150ml ] 2) A 20.5g marble chip containing CaCO 3 ( not 100% ) reacts with 2.0M 100ml HCl. CaCO 3 + 2HCl  CaCl 2 + H 2 O + CO 2

Exercise Calculate : Mole of HCl b) Mole of CaCO 3 c) Mass of CaCO 3 in the marble chip % by mass of CaCO 3 in the marble chip * [ 48.78% ]

Exercise A 1.306g tablet containing Vitamin C @ ascorbic acid was crushed & dissolved in water to form a solution. The solution was titrated with 0.050M I 2 solution. The end point is at 28.40ml.Determine % by mass of Vitamin C in the tablet. * [ 19.13% ] C 6 H 8 O 6 + I 2  C 6 H 6 O 6 + 2H + + 2I -

Redox Titration Example 1 5Fe 2+ + MnO 4 - + 8H +  5Fe 3+ + Mn 2+ + 4H 2 O Calculate molarity of 100ml FeSO 4 solution needed to react with 0.6M 50ml KMnO 4 solution.

Redox Titration Solution Method 1 M a V a / M b V b = a/b a = Fe 2+ , b = MnO 4 - M a x 100 / 0.6 x 50 = 5 / 1 M a (Fe 2+ or FeSO 4 ) = 1.5M @ mol/L

Redox Titration Method 2 Mole KMnO 4 @ MnO 4 - = 0.6 x 0.05 = 0.03 1 mole MnO 4 - requires 5 moles Fe 2+ Mole Fe 2+ or FeSO 4 = 0.03 x 5 = 0.15 Molarity of Fe 2+ or FeSO 4 = 0.15/0.1 = 1.5M

Redox Titration Example 2 The alcohol content of a brandy sample was found by taking 10ml sample & diluting to 500ml. 20ml of the solution requires 18.00ml 0.10M K 2 Cr 2 O 7 solution.Calculate molarity of ethanol , C 2 H 5 OH in the brandy sample. 3C 2 H 5 OH + 2Cr 2 O 7 2- + 16H +  4Cr 3+ 3CH 3 COOH + 11H 2 O

Redox Titration Solution n K 2 Cr 2 O 7 @ Cr 2 O 7 2- =0.018 x 0.1= 0.0018 n C 2 H 5 OH = 0.018 x 3/2 = 0.0027 (in 20ml) n C 2 H 5 OH (in 500ml) = 0.0027 x 500/20 = 0.0675 Molarity of C 2 H 5 OH = 0.0675 / (10/1000) ( in 10ml sample ) = 6.75M

Exercise Household bleach contains ClO - ion or NaOCl. The amount of ClO - in bleach can be determined by titrating ClO - with KI( I - ) to form chloride ion and iodine, I 2 . Construct an ionic (redox) equation based on the above statement. I 2 formed in a) reacts with S 2 O 3 2- to form I - and S 4 O 6 2- . Construct an ionic (redox) equation.

Exercise A 25.0ml sample of bleach is diluted to 250ml. 25ml solution portion is reacted with KI( I - ) to form I 2 . I 2 , iodine formed is then titrated with 20ml 0.2M Na 2 S 2 O 3 @ S 2 O 3 2- . Calculate : i) Mole of S 2 O 3 2- ii) Mole of I 2 iii) Mole of ClO - in the 25ml solution/portion iv) Thus, molarity of ClO - @ NaOCl in the bleach Answers = i) 0.004 ii) 0.002 iii) 0.002 iv) 0.8M

Solution 2H + + ClO - + 2I - Cl - + I 2 + H 2 O I 2 + 2 S 2 O 3 2- 2I - + S 4 O 6 2- I. n S 2 O 3 2- = 0.2mol L -1 x 20ml /1000mL L -1 = 0.004 mol iii. nI 2 = 0.004 mol/2 = 0.002 mol iii) n ClO- in 25ml sol’n/portion = nI 2 = 0.002 mol iv) In 25ml sol’n/portion: 25 x M ClO- /(20mL x 0.2 M S 2 O 3 2- )= 1/2 = 20mL x 0.2 M/25 x2 mL =0.080 M M ClO- in bleach = 250 x 0.080M /25 = 0.80 M

FURTHER EXERCISE: As ample of 1.55 g of iron ore is dissolved in an acid to give Fe 2+ solution. The solution formed is titrated with KMnO 4 to oxidize it to Fe 3+. The MnO 4 - ions are reduced to Mn 2+ . 92.95 mL of 0. 02 M KMnO 4 is required to reach the equivalent point of the titration. Write a balanced chemical equation for the reaction in the titration. Calculate the % of iron in the sample.

Solution i. 5Fe 2+ (aq) + MnO 4 - + 8 H + 5Fe 3+ (aq) + Mn 2+ + 4H 2 O Method 2 Mole KMnO 4 @ MnO 4 - = 0.02 x 0.09295 = 0.001859 1 mole MnO 4 - requires 5 moles Fe 2+ Mole Fe 2+ or FeSO 4 = 0.001859 x 5 = 0.009295 Mass of Fe in iron ore = 0.09295 x 56 = 0.521 g % Fe in sample = 33.56 %

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