STOICHIOMETRY 1. Balance the following redox reaction by ion- electron method taking place in acidic medium

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1 STOICHIOMETRY Short Answer Questions: 1. Balance the following redox reaction by ion- electron method taking place in acidic medium Cr O NO Cr NO. 7 Solution: Writing oxidation numbers Cr O N O Cr N O Locating atoms undergoing change in oxidation numbers Cr O N O Cr N O Dividing the reaction into two halves and balancing in acidic medium, separately Oxidation half-reaction: Reduction half-reaction N O N O Cr O 7 Cr Step1:Balance oxygen atoms N O H O N O Cr O Cr 7H O 7 Step: Balance hydrogen atoms in acidic medium N O H O N O H Cr O 1H Cr 7H O 7 Step: Balance the charge N O HO N O H e... (a) Cr O 7 1H 6e Cr 7HO (b)

2 Equalising the electrons and adding the two halves. eq (a) eq (b) 1, we get N O H O N O 6H 6e Cr O 1H 6e Cr 7H O 7 Cr O N O 8H Cr N O H O. This is the balanced equation. 7. Balance the following redox reaction by ion electron method taking place in acidic medium MnO SO Mn SO 1 Solution: Writing oxidation numbers 7 6 MnO S O Mn S O Locating atoms undergoing change in oxidation numbers 7 6 MnO S O Mn S O Dividing the reaction into two halves and balancing in acidic medium, separately Oxidation half-reaction: SO SO Reduction half-reaction: MnO Mn Step1:Balance oxygen atoms SO H O SO MnO Mn H O Step: Balance hydrogen atoms in acidic medium SO H O SO H 8 MnO H Mn H O Step: Balance the charge SO HO SO H e... (a) 8 5 MnO H e Mn H O... (b)

3 Equalising the electrons and adding the two halves. eq (a) 5 eq (b), we get 5SO 5H O 5SO 10H 10e MnO 16H 10e Mn 8H O MnO 5SO 6H Mn 5SO H O This is the balanced equation.. Iodate oxidises chromic hydroxide and gives iodide and chromate in basic medium. Solution: The ionic skeleton equation is written as OH ( ) IO Cr OH I CrO Writing oxidation numbers I O Cr( OH ) I Cr O Locating atoms undergoing change in oxidation numbers I O Cr( OH ) I Cr O Dividing the reaction into two halves and balancing in basic medium, separately Oxidation half-reaction: Reduction half-reaction: Cr( OH ) CrO IO I Step1: Balance oxygen atoms Cr( OH ) H O CrO IO I HO Step: Balance hydrogen atoms Cr( OH ) H O 5OH CrO 5H O IO 6HO I HO 6OH Step: Balance charge Cr( OH ) 5OH CrO HO e... (a) IO HO 6e I 6OH... (b)

4 Equalising the elements and adding the two halves eq (a) eq (b) 1, we get Cr( OH ) 10OH CrO 8H O 6e 6 6 IO H O e I OH IO Cr (OH) OH I CrO 5HO This is the balanced equation. -. White phosphorous reacts with aqueous caustic soda to give hypophosphite and phosphine. Solution: The ionic skeleton equation is OH P PH H PO Writing oxidation numbers P P H H PO Locating atoms undergoing change in oxidation numbers 0 1 P P H H PO d) Dividing the reaction into two halves and balancing in acidic medium, separately Oxidation half-reaction: Reduction half-reaction: P H PO P PH Step1: Balance phosphorous atoms P HPO P PH Step: Balance oxygen atoms P 8HO HPO P PH Step:Balance hydrogen atoms P 8HO 8OH HPO 8HO P 1HO PH 1OH Step: Balance charge P 8OH HPO e... ( a) P 1HO 1e PH 1OH (b)

5 Equalising the electrons and adding the two halves eq (a) eq (b) 1, we get P OH 1H PO 1e P 1H O 1e PH 1OH 1H O PH PO P 1OH 1H H H O PH P P O H O. This is the balanced equation. 5. A carbon compound contains 1.8% carbon,.1% hydrogen, 85.1% bromine. The molecular weight of the compound is Calculate the molecular formula. Solution: Step1: Percentage composition of the elements present in the compound. C H Br Step: Dividing with the respective atomic weights of the elements. 1.8/1= /1= /80= Step: Dividing by the smallest number to get simple atomic ratio /1.067= 1.1/1.067= 1.067/1.067= 1 The empirical formula is CHBr. Empirical formula weight 1 ( x 1) 80 = 9 The molecular weight = (given) n = = The molecular formula = (empirical formula) =( CH Br ) = C H Br

6 6. What are disproportination and Comproportination reaction? Give one example to each. Ans). The reactions in which the same element undergoing both oxidation and reduction simultaneously are called Disproportionation. Examples; a) Reaction of white phosphorous in aqueous caustic soda solution. PNaOHHO PHNaHPO b) Reaction of hot concentrated potash with bromine. 6KOH Br 5KBr KBrO HO Comproportionation: The reverse of disproportionation is comproportionation. In a comproportionation reaction, two species with the same element in two different oxidation states form single product. The element in the product is in an intermediate oxidation state, between that in reactants. Ex; Divalent silver oxidises metallic silver and it self is reduced to monovalent silver. AgSO Ag AgSO 7. Calculater the molarity of NaOH in a solution prepared by dissolving gm in enough water to form 50ml of the solution. weightofsolutex 1000 X1000 Solution: Molarity = 0.M GMWXvolumeofsolutioninml = 0X 50 =

7 Very Short Answer Questions 1. How many moles of glucose are present in 50 gm of glucose? Ans. Molar mass of glucose is 180gm Number of moles= weight/gmw= 50/180= moles.. Calculate the weight of 0.1 mole of Na CO? Ans. Weight of 0.1 mole of Na CO = moles GMW= 0.1x106=10.6 gm.how many glucose molecules are present in 5. gm of glucose (Molecular weight of glucose is 180U)? Ans. GMW i.e. 180 gm of Glucose= 6.0x10 molecules. 5. gm of glucose =? 5. No. of glucose molecules= X X = 1.75 x 10 molecules. Calculate the number of molecules present in 1.1 x 10 7 c.c. of a gas a STP? Solution: At STP, 00 cc = molecules At STP, 1.1 x 10 7 c.c =? Number of molcecules present in 1.1 x 10 7 c.c at STP = 1.1X X =.01 x Empirical formula of a compound is CH O molecular weight is 90, find molecular formula of that compound? Solution: Empirical formula of the compound = CH O Empirical formula weight = (1) ( x 1) 1(1 x 16) = 0 Molecular weight given = 90 Molecular weight 90 n = = = Empirical formula weight 0 Molecular formula =( Empirical formula)xn= (CH O) X = C H 6 O

8 Calculate the oxidation number of chromium in (i) Cr O 7 (ii) CrO - Ans. Oxidation number of chromium in (i) Cr O 7 = x 7(-) = - x=6 (ii) CrO - = x(-)= - x=6 7. Calculate the volume occupied by.5 Moles of a gas at STP? Ans. volume of 1mole of a gas at STP=. lit Volume of.5 moles of a gas at STP=.5 X. lit= 56 lit. 8. What volume of CO is obtained at STP by heating 10gms of CaCO? Ans. CaCO CaO CO 1 mole of CaCO gives 1 mole of CO i.e.. lit at STP i.e. 100 gms of CaCO gives. lit COat STP Volume of CO at STP given by 10 gms CaCO = X = lit 9. State (i) Law of definite proportions (ii) Law of multiple proportions Ans. (i) Law of definite proportions states that A given chemical substance always contains the same elements combined in a fixed proportion by weight. (ii) Law of multiple proportions states that If two elements chemically combine to give two or more compounds, then the weight of one element which combine with fixed weight of the other element in those compounds bear a simple multiple ratio to one another. 10. What is a red-ox reaction? Give an example? Ans. The reaction in which both oxidation and reduction takes place simultaneously is called a Red-Ox reaction. Ex; Zn CuSO Cu ZnSO In the above reaction zinc under goes Oxidation and copper ion under goes reduction.

GAS FORMULAE THE GENERAL GAS EQUATION. 1 dm = 1000 ml = 1 L. 1cm = 1 ml

GAS FORMULAE THE GENERAL GAS EQUATION PV = nrt Note Useful Conversions P = Pressure ( kpa ) 760 mmhg = 1 atmosphere V = Volume ( L ) = 101,35Pa = 101.35 kpa n = Amount (in mol) of gas ( mol) 1 1 R = Gas