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1 Chapter No. 1 BASIC CONCEPTS TEXT BOOK EXERCISE Q1. Select the most suitable answer from the given ones in each question. (i) The mass of one mole of electrons is (a) Properties which depend upon mass (b) Arrangement of electrons in orbital (c) Chemical properties (d) The extent to which they may be affected in electromagnetic field (ii) Which of the following statements is not true? (a) isotopes with even atomic masses are comparatively abundant (b) isotopes with odd atomic masses and even atomic number are comparatively abundant (c) atomic masses are average masses of isotopes. (d) Atomic masses are average masses of isotopes proportional to their relative abundance (iii) Many elements have fractional atomic masses, this is because (a) The mass of the atom is itself fractional (b) Atomic masses are average masses of isobars (c) Atomic masses are average masses of isotopes. (d) Atomic masses are average masses of isotopes proportional to their relative abundance (iv) The mass of one mole of electrons is (a) 008mg (b) 0.55mg (c) 0.184mg (d) 1.673mg (v) 27g of Al will react completely with how much mass of O 2 to produce Al 2 O 3 (a) 8g of oxygen (b) 16g of oxygen (c) 32g of oxygen (d) 24g of oxygen (vi) The number of moles of CO 2 which contain 8.0 g of oxygen. (a) 0.25 (b) 0.50 (c) 1.0 (d) 1.50

2 (vii) The largest number of molecules are present in (a) 3.6g of H 2 O (b) 4.8g of C 2 H 5 OH (c) 2.8 g of CO (d) 5.4g of N 2 O 5 (viii) One mole of SO 2 contains (a) 6.02x10 23 atoms of oxygen (b) 18.1x10 23 Molecules of SO 2 (c) 6.02x1023 atoms of sulphur (d) 4 gram atoms of SO 2 (ix) The volume occupied by 1.4 g of N 2 at STP is (a) 2.24 dm 3 (b) 22.4dm 3 (c) 1.12 dm 3 (d) 112 cm 3 (x) A limiting reactant is the one which (a) is taken in lesser quantity in grams as compared to other reactants (b) is taken in lesser quantity in volume as compared to the other reactants (c) give the maximum amount of the product which is required (d) give the minimum amount of the product under consideration Ans: (i)a (ii)d (iii)d (iv)b (v)d (vi)a (vii)a (viii)c (ix)c (x)d Q2: Fill in the blanks: (i) The unit of relative atomic mass is (ii) The exact masses of isotopes can be determined by spectrograph. (iii) The phenomenon of isotopes was first discovered by (iv) Empirical formula can be determined by combustion analysis for those compounds which have and in them. (v) A limiting reagent is that which controls the quantities of (vi) I mole of glucose has atoms of carbon of oxygen and of hydrogen. (vii) 4g of CH 4 at O o C and I atm pressure has molecules of CH 4. (viii) Stoichiometry calculations can by performed only when law is obeyed. Ans: (i) amu (ii) mass (iii) Soddy (iv) carbon, hydrogen (v) Products (vi) 6N A,6N A,12N A (vii) 1.505x10 23 (viii) conservation and multiple proportion Q3: Indicate true or false as the case may be: (i) (ii) Neon has three isotopes and the fourth one with atomic mass amu. Empirical formula gives the information about he total number of atoms present in the molecule

3 (iii) During combustion analysis Mg(CIO 4 ) 2 is employed to absorb water vapors. (iv) Molecular formula is the integral multiple of empirical formula and the integral multiple can never be unity. (v) The number of atoms in 1.79 g of gold and 0.023g of sodium are equal. (vi) The number of electrons in the molecules of CO an dn 2 are 14 each, so 1 mg go each gas will have same number of electrons. (vii) Avogadro s hypothesis is applicable to all types of gases, i.e., ideal and non-ideal. (viii) Actual yield of a chemical reaction may by greater than the theoretical yield. Ans. (i) False (ii) False (iii) True (iv) False (v) False (vi) True (vii) False (viii) False Q4: What are ions? Under What condition are they produced? Ans: Ions can be produced by the following processes: (i) By dissolving ionic compounds in water (ii) By X-rays (iii) In mass spectrometry (iv) By removing or adding electron in atom Q4: (a) (b) (c) What are isotopes? How do you deduce the fractional atomic masses of elements form the relative isotopes abundance? Give two examples in support of your answer. ( See detail in Sublime subjective) How does a mass spectrograph show the relative aboundace of isotopes of an element?. ( See detail in Sublime subjective) What is the justification of two strong peaks in the mass spectrum for bromine; while for iodine only one peak at 127 amu, is indicated? Ans The two strong peak in the mass spectrum for bromine represent two different isotopes of bromine having nearly equal natural abundances. Only one peak at 127 amu in the mass spectrum for iodine indicates that it has only one isotope of atomic mass 127 amu. Remember that! Height of the peaks Relative abundance of isotopes No. of peaks = No. of isotopes Q5: Silver has atomic number 47 and has 16 known isotopes but two occur naturally I,e, Ag 107. and Ag 109. Given the following mass spectrometric data, calculated the average atomic mass of silver,

4 Isotopes mass (amu) percentage abundance 107 Ag Ag The mass contribution for silver are: Isotopes Fractional abundance isotopic mass mass contribution 107 Ag x107= Ag x109= Fractional atomic mass of silver = Hence the fractional atomic mass of silver is = Ans. Q6: Boron with atomic number 5 has two naturally occurring isotopes. Calculate the percentage abundance of 10 B and 11 B from the following information. Average atomic mass of boron =10.81 amu Isotopic mass of 10 B = amu Isotopic mass of 11 B = Let, the fractional abundance of 10 B =x The fractional abundance of 11 B =1-x Remember that the sum of the fractional abundances of isotopes must be equal to one, now, The equation to determine the atomic mass of element is (fractional abundance) (isotopic mass) (fractional abundance of 10 B)(isotopic mass of 10 B )+(fractional abundance of 11 B) (isotopic mass of 11 B) =Average atomic mass of Boron (x)( )+(1-x)( ) = x x = x x = x = x = Fractional abundance of 10 B = Fractional abundance of 11 B =( )= By percentage the fractional abundance of isotope is %of 10 B =0.2000x100 =20% Answer % of 11 B =0.8000x100 =80%Answer Q7: Define the following terms and give three examples of each. (i) Gram atom (ii) Gram molecular mass

5 (iii) Gram molecular mass (iv) Gram ion (v) molar volume (vi) Avogadro s number (vii) Stoichiometry (viii) Percentage yield Q8: Justify the following statements: (a) 23 g of sodium and 238g of uranium have equal number of atoms in the (b) Mg atom is twice heavier than that of carbon (c) 180g of glucose and 342 g of sucrose have the same number of molecules but different number of atoms present in them. (d) 4.9g of H 2 SO 4 when completely ionized in water, have equal number of positive and negative charges but the number of positively charged ions are twice the number of negatively charged ions. (e) One mg of K 2 C r O 4 has thrice the number of ions than the number of (f) formula units when ionized in water. Two grams of H 2, 16 g of ch4 and 44g of CO 2 occupy separately the volumes of dm 3, although the sizes and masses of molecules of three gases are very different from each other. (a) 23g of Na =1 mole of Na =6.02x10 23 atoms of Na 238g of U =1 mole of U =6.02x10 23 atoms of U. Since equal number of gram atoms(moles) of different elements contain equal number of atoms. Hence, 1 mole (23g ) of sodium and 1 mole (238)g of uranium contain equal number of atoms, i, e,6.02x10 23 atoms. (b) Since the atomic mass of Mg (24) is twice the atomic mass of carbon (12) therefore, Mg atom is twice heavier than that of carbon. Or Mass of 1 atom of Mg= Mass of 1 atom of C = Since the mass of one atom of Mg is twice the mass of one atom of C, therefore, Mg atom is twice heavier than that of carbon. (c) 180 g of glucose = 1 mole of glucose =6.02x10 23 molecules of glucose 342 g of sucrose=1mole of sucrose =6.02x10 23 molecules of sucrose Since one mole of different compounds has the same number of molecules. Therefore 1 mole (180g) of glucose and I mole (342g) of sucrose contain the same number (6.02x10 23 )of molecules. Because one molecule of glucose, C 6 H 12 O 6 contains 45 atoms whereas one molecules of glucose, C 12 H 22 O 11 contains 24 atoms. Therefore, 6.02x10 23 molecules of glucose contain different atoms as compound to6.02x10 23 molecules of sucrose. Hence, 180 g of glucose and 342g og sucrose have the same number of molecules but different number of atoms present in them. (d) H 2 SO 4 2H + + SO

6 When one molecules of H 2 SO 4 completely ionizes in water it produces two H + ion and one SO ion,. Hydrogen ion carries a unit positive charge whereas SO ion carries a double negative charge. To keep the neutrality, the number of hydrogen are twice than the number of soleplate ions. Similarly the ions produced by complete ionization of 4.8g of H 2 SO 4 in water will have equal number of positive and negative but the number of positively charged ions are twice the number of negatively charged ions. (e) H 2 SO 4 2H + + SO K 2 C r O 4 when ionizes in water produces two k + ions one C O ion. Thus each formula unit of K 2 C r O 4 produces three ions in solution.hence one mg of K 2 C r O 4 has thrice the number of ion than the number of formula units ionized in water. (f) 2g of H 2 =1 mole of H 2 =6.02x10 23 molecules of H 2 at STP =22.414dm 3 16g of CH 4 =1mole of CH 4 =6.02x10 23 molecules of CH 4 at STP =22.414dm 3 144g of CO 2 =1mole of CO 2 =6.02x10 23 molecules of CO 2 at STP =22.144dm 3 Although H 2, CH 4 and CO 2 have different masses but they have the same number of moles and molecules. Hence the same number of moles or the same number of molecules of different gases occupy the same volume at STP. Hence 2 g of H 2,16g of CH 4 and 44 g of CO 2 occupy the same volume dm 3 at STP. The masses and the sizes of the molecules do not affect the volumes. Q10: Calculate each of the following quantities (a) Mass in grams of 2.74 moles of KMnO 4. (b) Moles of O atoms in 9.0g of Mg (NO 3 ) 2. (c) Number of O atoms in g of Cu SO 4.5H 2 O. (d) Mass in kilograms of 2.6x molecules of SO 2. (e) Moles of C1 atoms in 0.822g C 2 H 4 C1 2. (f) Mass in grams of moles of silver carbonate. (g) Mass in grams of 2.78x10 21 molecules of CrO 2 C1 2. (h) Number of moles and formula units in 100g of KC1O 3. (i) Number of K + ions C1O ions, C1 atoms, and O atoms in (h) (a) No of moles of KMnO 4 =2.74moles formula mass of KMnO 4 = =158g mol -1 Mass of KMnO 4 =? Formula used: Mass of KMnO 4 = no of mole of KMnO 4 x formula mass of KMnO 4 =2.74 mol x 158 g mol-1 =432.92g Answer

7 (b) Mass of Mg (NO 3 ) 2 =9g Formula mass of Mg (NO 3 ) 2 = =148g mol -1 No of moles of O atoms =? Formula used: No of mole of Mg (NO 3 ) 2 = Now, I mole of Mg (NO 3 ) 2 contains = 6moles of O atoms moles of Mg (NO3)2contains =6x0.6 =0.36 moles of O atoms Alternatively, 148g of Mg (NO 3 ) 2 contains =6moles of O atoms g of Mg (NO 3 ) 2 contains = =0.36 mole Answer (c) Mass of CuSO 4. 5H 2 O=10.037g Formula mass of CuSO 4. 5H 2 O= = g mol -1 No of moles of CuSO 4. 5H 2 O =? No of moles of CuSO 4. 5H 2 O = = Now, 1 mole of CuSO 4.5H 2 O contains 9moles of O atoms 0.04 mole of CuSO 4.5H 2 O contains=9x0.04 =0.36 moles of O atoms Now, I mole of O atoms contains =6.02x10 23 O atoms 0.36 mole of O atoms contains =6.02x10 23 x0.36 oxygen atoms =2.17x10 23 oxygen atoms =2.17x10 23 atoms Answer (d) No of molecules of SO 2. =2.6x10 20 molecules Molecular mass of SO 2. =32+32=64 g mol -1 Now, Avogadro s number, N A =6.02x10 23 molecules of SO 2 Mass of SO 2 molecules =27.64x10-3 g = =27.64x10-6 kg =2.764x10-3 kg Answer (e) Mass of C 2 H 4 C1 = 0.822g

8 Molecular mass of C 2 H 4 C1 = =99 g mol-1 No of moles of C 2 H 4 C1 = Now, 1 mole of C 2 H 4 C1 contains =2moles of C1 atoms 8.3x10-3mole of C 2 H 4 C1 contains =2x8.3x10-3 mole of atom =16.6x10-3 =0.0166mole of C1 atom =0.017 mole Answer (f) No of mole of Ag 2 CO 3 =5.136moles Formula mass of Ag 2 CO 3 = = g mol -1 Mass of Ag 2 CO 3 =No of moles of Ag 2 CO 3 xformula mass of Ag 2 CO 3 =5.136molx g mol -1 =416.18g = g Answer (g) Molecular mass of CrO 2 C1 2 = =155g mol -1 N A =6.02x10 23 molecules mol -1 Molecules of CrO 2 C1 2= =2.78x10 21 molecules Now, mass of CrO 2 C1 2 = = =71.578x10-2 g = =0.716 g Answer (h) Mass of KCIO 3 =100g Formula mass of KCIO 3 =39x =122g mol -1 No of moles of KCIO 3 =? No of moles of KCIO 3 = = =0.816mole Answer No of formula units No of moles x Avogadro,s No =0.816mole x 6.02x10 23 formula units =4.91x10 23 formula units (i) No of K + ions =4.91x10 23 Answer No of CIO ions =4.91x10 23 Answer No of CIO ions =4.91x10 23 Answer No of O atoms = 4.91x10 23 x3 =14.73x10 23 =1.473x10 24 Answer Q 11 Aspartame he artificial sweetener, has a molecular formula of C 14 H 18 N 2 O 5. (a) What is the mass of one mole of aspartame?

9 (b) How many moles are present in 52g of aspartame? (c) What is the mass in grams of moles of aspartame? (d) How many hydrogen atoms are present in 2.34g of aspartame? (a) Molecular mass of aspartame = =295g mol -1 Mass of 1 mole of aspartame =294g mol -1 Answer (b) Mass of aspartame =52g Molecular mass of aspartame =294g mol -1 No of moles of aspartame = = = mol =0.177 mol Answer (c) No moles of aspartame = moles Molecular mass of aspartame =294g mol -1 Mass of aspartame =No of moles x Molar mass =10.122mol x 294g mol -1 = g Answer (d) Mass of aspartame =243g Molar mass of aspartame =294g mol -1 No of molecules of aspartame=? No of molecules of aspartame= xn A = = =4.98x10 21 molecules. Now,1 molecule of aspartame contains =18 H atoms 4.98x 1022 molecules =18x4.98x10 21 H atoms =89.64x10 21 H atoms =8.964x1022 H atoms Answer Q 12: A sample of mole of a metal M reacts completely with excess of fluorine to from 46.8g MF 2. (a) How many moles of F are present in the sample of MF 2 that forms. (b) which elements is represented by the symbol M? (a) Formula of compound =MF 2 No of moles of M =0.6 mol Mass of MF 2 =46.8g The molar of M:F in the compounds;

10 No of moles of F =0.6x2=1.2mol Answer Mass of F =No of moles of Fx At. mass of F =1.2x19=22.8g Mass of compound =46.8g Mass of metal, M = =24 At mass of M = = (b) The atomic mass of the elements, M =40 The metal is calcium, Ca Answer Q 13 : In each pair, choose the larger of the indicated quantity,or state if the samples are equal. (a) Individual particles: 0.4 mole of oxygen molecules or0.4mole of oxygen atom. (b) Mass: 0.4 mole of ozone molecules or0.4mole of oxygen atoms (c) Mass: 0.6 mole of C 2 H 4 or 0.6mole of 1 2 (d) Individual particles: 4.0g N 2 O 4 or 3.3g SO 2 (e) Total ions: 2.3 moles of NaC1O 3 or 2.0mole of MgC1 2 (f) Molecules: 11.0g of H 2 Oor 11.0g H 2 O 2 (g) Na + ion: moles of NaBr or kg NaC1 (h) Mass: 6.02x10 23 atoms of 235 U or 6.02x1023 atoms of 238 U Ans: (a) Number of molecules =moles x N A Number of O 2 molecules =0.4x6.02x10 23 =2.408x10 23 molecules No of O atoms=0.4x6.02x10 23 =2.108x10 23 atoms There are equal number of individual particles in 0.4 mole of oxygen molecules and 0.4 mole of oxygen atom. In general, equal number of moles of different substances contains equal number of particles. Both are equal Answer (b) Mass of substance = moles x molar mass Mass of oxygen atoms =0.4x16=64g Mass of ozone, O 3 molecules =0.4x48=19.2g

11 0.4 moles of ozone molecules have larger mass than 0.4mole of oxygen atoms. Ozone Answer (c) Mass of C 2 H 4 =0.6x28=1.68g Mass of 1 2 =0.6x127=254g 0.6mole of 12 have larger mass than 0.6 mole of C 2 H Answers (d) No of molecules = No of molecules in N 2 O 4 = x6.02x10 23 =2.62 x10 23 molecules No of molecules in SO 2 =x6.02x10 23 =3.1x10 22 molecules 3.3g of SO 2 have larger number of individual particles than 4.0 g of N 2 O 4. SO 2 Answer (e) No of formula units =Moles x N A No of formula units of NaC1O 3 =2.3x6.02x10 23 =1.38x10 24 formula units No of ions in 1 formula units of NaC1O 3 =2 Total no of ions in MgC1 2 =2x1.38x10 23 =2.76x10 24 ions No of formula units of MgC1 2 =2.0x6.02x ions x3=3.6x10 No.of ions in one formula unit of MgC1 2 =3 Total no of ions in MgC1 2 =1.20x10 24 x3=3.6x10 24 ions 2.0moles of MgC1 2 contain lager number of total ions than 2.3 moles of NaC1o 3- MgC1 Answer (f) No of molecules = N A No of molecules in H 2 O 2 = x6.02x10 23 =3.68x10 23 molecules No of molecules in H 2 O 2 = x6.02x10 23 =1.95x10 23 molecules 11.0g of H 2 O 2 contains larger number of molecules than 11.0g of H 2 O 2 H 2 O 2 Answer (g) No of formula units =moles xn A No of formula units NaBr =0.5x6.02x10 23 =3.01x10 23 formula units One formula units o NaBr contain Na + ions = x10 23 formula unit of NaBr contains Na + ions =3.01x10 23 Na + ions units No of formula units of NaC1 = x6.02x10 23 =1.49x10 23 formula One formula unit of NaC1 contains Na + ions =1 1.49x10 23 formula units of NaC1 contains =1.49x10 23 Na + ions

12 0.500 moles of NaBr contains lager number of Na + ions than kg ofnac1. NaBr Answer (h) Mass of atoms of an element = Mass of 235 Uatoms =x6.02x10 23 =235g Mass of 238 U atoms =x6.02x10 23 =238g 238 U Answer Q 13: (a) (b) Calculate the percentage of nitrogen in the four important fertilizer i.e., (i)nh 3 (ii)nh 2 CONH 2 (Urea) (iii)(nh 4 ) 2 SO 4 (iv)nh 4 NO 3 Calculate the percentage of nitrogen and phosphorus in each of the following: (i) NH 4 H 2 PO 4 (ii) (NH 4) ) PO 4 (iii) (NH 4 ) 4 PO 4 (a) Mol-mass of NH 3 =14+4=17g Mass of N =14g % of N =x100 =82.35% Answer (b) Mol-mass of NH 2 CONH 2 = =60g Mass of N =28g %of N =x100 =46.35% Answer (c) Mol-mass of (NH 2 ) 2 SO 4 = =132g Mass of N =28g % of N =x100 =21.21% Answer (d) Mol-mass of (NH 2 ) 2 SO 4 = =80g Mass of N =28g %of N =x100 =35% Answer (I) Mol-mass of (NH 2 ) 2 SO 4 = =115g Mass of N =14g Mass of P =31g %of N =x100=12.17% Answer %of P ==26.96% Answer (II) Mol-mass of ((NH 2 ) 2 SO 4 = =132g Mass of N =28g Mass of P = %of N = =21.21% Answer %of P = =23.48% Answer (III) Mol-mass of (NH 2 ) 2 SO 4 = =149g

13 Mass of N =42g Mass of P =31g %of N = %of P = Q 14: Glucose C 6 H 12 O 6 is the most important nutrient in the cell for generating chemical potential energy. Calculate the mass% of each element in glucose and determine the number of C,H and O atoms in 10.5g go the sample. Mol-mass of glucose C 6 H 12 O 6 = =180g Mass of C =72 Mass of H =12 Mass of O =96 % of C = =40% Answer % of H = =6.66% Answer % of O = =53.33% Answer Mass of C 6 H 12 O 6 =10.5g Mol-mass of C 6 H 12 O 6 =180g Mol-mass of =180g mol -1 No of moles of C 6 H 12 O 6 = No of molecules of glucose =No of moles x N A =0.058 molx 6.02x10 23 molecules mol -1 =0.35x10 23 molecules =3.5x10 22 molecules Now, 1 molecule of glucose contains =6C-atoms 3.4x10 22 molecules of glucose contains =6x3.5x10 22 C-atoms =21x10 22 =2.1x10 23 C atoms Answer 1 molecules of glucose contains =12H-atoms 3.5x10 22 molecules glucose contains =12x3.5x10 22 =4.2x10 23 H- atoms Answer 1 molecule of glucose contains =6 O atoms 3.5 x molecules of glucose contains =6x3.5x10 22 =2.1x10 23 O-atoms Answer Q 16: Ethylene glycol is used as automobile antifreeze.it has 38.7% carbon, 9.7% hydrogen and 51.6% oxygen. Its molar mass is 62.1 grams mol -1.Determine its molecular formula. % of C=38.37 g % of H =9.7g % of O=51.6g At. Mass of C=12g mol -1 At. Mass of H=1.008g mol -1 At. Mass of O =16g mol -1

14 No of moles of C = No of moles of H = No of moles of O = Atomic ratio is obtained by dividing the moles with 3.23, which is the smallest ratio. C :H :O 1 :3 :1 Empirical formula =CH 3 O Empirical formula mass =31 n= Molecular formula =2x CH 3 O =C 2 H 6 O 2 Answer Q 16: Serotonin (Molecular mass= 176g mol -1 ) is a compound that conducts nerve impulses in brain and muscles. It contains 68.2 % C, 6.86% H, and 9.08% O. What is its molecular formula? No of moles of C = No of moles of H = No of moles of N = No of moles of O = Atomic Ratio C : H : N : O : : : 10 : 12 : 2 : 1 Empirical formula =C 10 H 12 N 2 O Empirical formula mass = =176g mol -1 Molecular mass =176g mol -1 n= Q17: An unknown metal M reacts with S to from a compound with a formula M 2 S 3.If 3.12 g of M reacts with exactly 2.88 g of sulphur, what are the names of metal M and the compound M 2 S 3. Formula of compound = M 2 S 3 Mass of M =3.12g Mass of S =2.88g

15 Atomic mass of S =32g mol -1 No of moles of S = No of moles of S = The molar ratio of M: S in the compound is : No of moles of M = =0.06 mole Now, No of moles of M = At. Mass M = The mass of M used in the formation of M 2 S 3 is 3.12g. The product M 2 S 3 therefore also contains 3.12g of M, because mass is conserved. The amount of M before and after reaction must be the same. Since we know both the number of moles of M and the mass of M, we can cal calculate the atomic mass of M as follows: At. Mass of M = =52 Atomic number, Z =52 Q19: The octane present in gasoline burns according to the following equation. 2C 8 H 18 (i) (g) 16CO 2(g) + 18H 2 O (i) (a) How many moles of O 2 are needed to react fully with 4 moles of actane? (b) How many moles of CO 2 can be produced from one mole of actane? (c) How many moles of water are produced by the combustion of 6 moles of octane? (d) If this reaction is to be used to synthesize 8 moles of CO 2 how many grams of oxygen are needed? How many grams of octane will be used? (a) 4 moles 2C 8 H 18 (i) (g) 16CO 2(g) + 18H 2 O (i) 2 moles 25 moles 2 moles of C 8 H 18 =25 moles of O2 4 moles of C 8 H 18 =

16 (b) =50moles of O 2 Answer 1 moles 2C 8 H 18 (i) (g) 16CO 2(g) + 18H 2 O (i) 2 moles Now, 2 moles of C 8 H 18 =16 moles of CO 2 1 mole of C 8 H 18 = =8 moles of CO 2 Answer (c) 6 moles 2C 8 H 18 (i) (g) 16CO 2(g) + 18H 2 O (i) 2 moles Now, 2 moles of C 8 H 18 =18 moles of H2 O(i) 6 moles of C 8 H 18 = =54 moles of H 2 O (d) 6 moles 2C 8 H 18 (i) (g) 16CO 2(g) + 18H 2 O (i) 2 moles 1800moles Now, 16 moles of CO 2 =25 moles of O 2 8 moles of CO 2 = =12.5 moles of CO 2 Mol-mass of O 2 =32g mol -1 =12.5 molx 32g mol -1 =400g of O 2 Now, 16moles of CO 2 =2moles of C 8 H 18 8 moles of CO 2 = =1 mole of C 8 H 18 Mol-mass of C 8 H 18 =96+18=114g mol-1 Mass of C 8 H 18 =No of moles of C 8 H 18 xmol.mass ofc 8 H 18 =1 molx 114 g mol g Answer Q19: Calculate the number of grams of A1 2 S 3 which can be prepared by the reaction of 20 g of A1 and 30 g of sulphur. How much the non-limiting reaction is in excess? 0.74 mole 0.94 mole Mass of A1 =20g Molar mass of A1 =27g mol -1 No of moles of A1 = Mass of S = 30g Molar mass of S =32g mol -1 No of moles of S =

17 2A1 + 3S A1 2 S 3 2 mole 3 mole 1 mole Now, 2 moles of A1 =1 mole of A1 2 S moles of A1 = =0.37 mole of A1 2 S 3 Now, 3 moles of S =1 moles of A1 2 S moles of S = =0.313 mole of A1 2 S 3 Since S give the least number of moles of A1 2 S 3 therefore, it is the limiting reactant. No of moles of A1 2 S 3 =0.313 mole Molar mass of A1 2 S 3 =150g mol-1 Mass of A1 2 S 3 =No of moles of A1 2 S 3 xmolar mass of A1 2 S 3 =0.313molx 150 g mol -1 =46.95 g of A1 2 S 3 Answer The non-limiting reactant is A1 which is in excess. Now mass of A1 required reacting completely with 0.94 moles of S can be calculated as: 0.94 mole 2A1 + 3S A1 2 S 3 2 mole 3 mole Now, 3 moles of S =2 moles of A moles of S = = Mass of A1 =No of moles of A1 x molar mass of A1 =0.63x 27 =17g of A1 Mass of A1available =20g Mass of A1 which reacts completely =17g with available S Excess of A1 =20-17=3g Q20: A mixture of two liquids, hydrazine N 2 H 4 and N 2 O 4 are used as a fuel in rockets. They produce N 2 and water vapors. How many grams of N 2 gas will be formed by reacting 100g of N 2 O 4 and 200g g of N 2 O 4. 2N 2 H 4 + N 2 O 2 3N 2 +4 H 2 O Mass of2n 2 H 4 =100g Mass of N 2 O 2 =200g Molar mass of 2N 2 H 4 =28+4=32g mol -1 Molar mass of N 2 O 2 =28+64=92g mol -1 No of moles of N 2 H 4 = No of moles of N 2 O 2 =

18 3.125moles moles 2N 2 H 4 + N 2 O 2 3N 2 +4 H 2 O 2 moles 1mole 3moles Now, 2moles of N 2 H 4 =3moles of N moles of N 2 H 4 = =4.69 mole of N 2 Now, 1 mole of N 2 O 2 =3moles of N moles of N 2 O 4 = =6.52 mole of N 2 O 2 Since N 2 H 4 gives the least number of moles of N 2, hence it is the limiting reactant. Amount of N 2 produced =4.69 moles Molar mass of N 2 =28g mol -1 Mass of N 2 =4.69g molx 28g mol-1 = g Answer Q21: Silicon carbide (SiC) is an important ceramic material. It is produced by allowing sand (SiO 2 )to react with carbon at high temperature. SiO 2 + 3C SiC + 2CO When 100kg sand isn reacted with excess of carbon, 51.4 kg of Sic is produced. Mass of SiO2 =100 kg=100000g Mass of SiC produced =5.14 kg =51400g g SiO 2 + 3C SiC + 2CO 60g 40g Now, 60g of SiO 2 =40g of SiC g of SiO 2 = = g Actual yield of Sic =51400 g Theoretical yield of SiC = g % yield = = =77.1% Q22: (a) What is Stoichiometry? Give its assumptions? Mention two important law, which help to perform the Stoichiometry calculations. (b) What is a limiting reactant? How does it control the quantity of the product formed? Explain with three examples Q 23: (a) Define yield. How do we calculate the percentage yield of a chemical reaction? (b) What are the factors which are mostly responsible for the low yield of the products in chemical reactions.

19 Q24: Explain the following with reasons. (j) Law of conservation of mass has to be obeyed during Stoichiometric calculations. (ii) Many chemical reactions taking place in our surrounding involves the limit reactants. (iii) No individual neon atom in the sample of the element has a mass of 20.18amu. (iv) One mole of H 2 SO 4 should completely react with two moles of NaOH. How does Avogadro, s number help to explain it. (v) One mole H 2 O has two moles of bonds, three moles of atoms, ten moles of electrons and twenty eight moles of the total fundamental particles present in it. (vi) N2 and CO have the same number of electrons, protons and neutrons. Ans. (i) According to law of conservation of mass, the amount of each element is conserved in a chemical reaction. Chemical equations are written and balanced on the basis of law of conversation of mass. Stoichiometry calculations are related with the amounts of reactants and products in a balanced chemical equation. Hence, law of conservation of mass has to be obeyed during stoichiometric calculations. (ii) In our surrounding many chemical reactions are taking place which involve oxygen. In these reactions oxygen in always in excess quantity while other reactant are in lesser amount. Thus other reactants act as limiting reactants. (iii) Since the overall atomic mass of neon in the average of the determined atomic masses of individual isotopes present in the sample of isotopic mixture.hence, no individual neon atom in the sample has a mass of 20.18amu. (iv) H 2 SO 4 +2NaOH Na 2 SO 4 + 2H 2 O 1 mole 2moles (v) 2 moles of H + ions 2 moles of OH ions 2x6.02x10 23 H + ions 2x6.02x10 23 OH ions Once mole of H 2 SO 4 consists of 2 moles of H + ions that contains twice the Avogadro s number of H + ions. For complete neutralization it needs 2 moles of one mole of H 2 SO 4 should completely react with two moles of N A OH. Since one molecule of H 2 O has two covalent bonds between H and O atoms. Three atoms, ten electrons and twenty eight total fundamental particles present in it. Hence, one mole of H 2 O has two moles of bond, three moles of atoms, ten moles of electrons and twenty eight moles of total fundamental particle present in it.

20 In N 2 there are 2 N atoms which contain 14 electrons (2x7),14 protons (2x7) and 14 neutrons (2x7). In CO, there are one carbon and one oxygen atoms. It contains 14 electrons (6carbon e +8 oxygen e), 14 protons (6 C proton +8 O proton ) and 14 neutrons (6 neutrons +8 O neutrons).hence, N 2 and CO have the same number of electrons, protons and neutrons. Remember that electrons, protons and neutrons of atoms remain conserved during the formation of molecules in a chemical reaction.