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1.1.9 Moles and reactions Deduce the quantities of reactants and products from balanced equations.

1.1.9 Moles and reactions Deduce the quantities of reactants and products from balanced equations. MOLES = MASS MOLAR MASS. THE MOLE. CALCULATING THE NUMBER OF MOLES OF A SINGLE SUBSTANCE moles = mass / molar mass mass = moles x molar mass

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1.1.9 Moles and reactions Deduce the quantities of reactants and products from balanced equations.

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  1. 1.1.9 Moles and reactions Deduce the quantities of reactants and products from balanced equations.

  2. MOLES = MASS MOLAR MASS THE MOLE CALCULATING THE NUMBER OF MOLES OF A SINGLE SUBSTANCE moles = mass / molar mass mass = moles x molar mass molar mass = mass / moles UNITS mass g or kg molar mass g mol-1or kg mol-1 MASS MOLES x MOLAR MASS COVER UP THE VALUE YOU WANT AND THE METHOD OF CALCULATION IS REVEALED

  3. ? grms 20.0 grms 2. Hydrochloric acid + Potassium carbonate → Potassium chloride + Water + Carbon dioxide 2HCl(aq) + K2CO3(aq) → 2KCl(aq) + H2O(l) + CO2(g) 3. ethanol     +     ethanoic acid        ethyl ethanoate   +   water.CH3CH2OH(aq)    +     CH3CO2H(aq)          CH3CH2CO2CH3(aq)   +   H2O(l) ? grms ? grms 1.0 grms ? tonnes 100 tonnes 1.0 tonnes = 1000 kg 1 kg = 1000 grms

  4. ? grms 20.0 grms 2. Hydrochloric acid + Potassium carbonate → Potassium chloride + Water + Carbon dioxide 2HCl(aq) + K2CO3(aq) → 2KCl(aq) + H2O(l) + CO2(g) 2x39.1+12.0+3x16.0= 138.2 Mr K2CO3 20.0 / 138.2 Moles K2CO3 …… Moles KCl Mass KCl 2x39.1+35.5=

  5. 3. ethanol     +     ethanoic acid        ethyl ethanoate   +   water.CH3CH2OH(aq)    +     CH3CO2H(aq)          CH3CH2CO2CH3(aq)   +   H2O(l) 2x39.1+12.0+3x16.0= 138.2 2x39.1+35.5= 138.2 138.2 ? grms ? grms 1.0 grms ? tonnes 100 tonnes 1.0 tonnes = 1000 kg 1 kg = 1000 grms

  6. 1.1.9 Tasks Write out these Key definitions Stoichiometry is . . . Examiner tip is . . . . Notes Worksheets Questions : 1, 2, 3

  7. 1.1.9 Moles and reactions Deduce the quantities of reactants and products from balanced equations.

  8. MOLES = MASS MOLAR MASS THE MOLE CALCULATING THE NUMBER OF MOLES OF A SINGLE SUBSTANCE moles = mass / molar mass mass = moles x molar mass molar mass = mass / moles UNITS mass g or kg molar mass g mol-1or kg mol-1 MASS MOLES x MOLAR MASS COVER UP THE VALUE YOU WANT AND THE METHOD OF CALCULATION IS REVEALED

  9. a) 1.51 x 1022 b) 3.42 x 1022 c) 1.45 x 1022 d) 2.15 x 1024 • a) 11.6 g b) 110 g c) 8.07 x 10-3 or 8.07 mg

  10. Reacting Masses 6. Calculate the mass of H2O required to react completely with 5.0 g of SiCl4: SiCl4 + 2H2O  SiO2 + 4HCl Moles SiCl4 = 5.0/ 170.1 = 0.02939 Ratio SiCl4 : H2O 1 : 2 Moles : H2O = 0.05879 Mass : H2O = 0.05879 x 18.0 = 1.0582 =1.06 grm Si : 28.1 x 1=28.1 Cl :35.5x4=142.0 170.1grm

  11. 7. Calculate the mass of phosphorus required to make 200 g of phosphine, PH3, by the reaction: NOW CORRECTED ! P4(s) + 4NaOH(aq) +2H2O(l)  2Na2HPO3(aq) + 2PH3(g) Moles PH3= 200 / 34.0 = 5.8823 Ratio P4 : PH3 1 : 2 Moles : P4 = 5.8823 / 2 = 2.94115 Mass : P4 = 2.94115 x 31.0 x 4 = 364.7026 grns = 365 grms 31.0+3.0=34.0

  12. 8. Lead (IV) oxide reacts with concentrated hydrochloric acid as follows: PbO2(s) + 4HCl(aq)  PbCl2(s) + Cl2(g) + 2H2O(l) What mass of lead chloride would be obtained from 37.2g of PbO2, and what mass of chlorine gas would be produced? Moles PbO2= 37.2 / 239.2 = 0.15551 Ratio PbO2 : PbCl2 1 : 1 Moles : PbCl2 = 0.15551 Mass : PbCl2= 0.15551 x 278.2 = 43.3 grms Ratio PbO2 : Cl2 1 : 1 Moles : Cl2 = 0.15551 mass : Cl2= 0.15551 x 71.0 = 11.01 grms 207.2+32.0 = 239.2 207.2+71.0= 278.2

  13. 9. When copper (II) nitrate is heated, it decomposes according to the following equation: 2Cu(NO3)2(s)  2CuO(s) + 4NO2(g) + O2(g). When 20.0g of copper (II) nitrate is heated, what mass of copper (II) oxide would be produced? What mass of NO2 would be produced? Moles Cu(NO3)2= 635.+[(14.0+16x3)x2]=187.5 20.0 / 187.5 = 0.10666 moles ( 0.107 or 1.07 x10-1 moles) Ratio Cu(NO3)2 : CuO 1 : 1 Moles : CuO = 0.10666 63.5+16.0=79.5 Mass : CuO= 0.10666 x 79.5 = 8.47947( 8.48 grms ) Ratio Cu(NO3)2 : NO2 2 : 4 Moles : NO2 = 0.10666x2= 0.21332 mass : NO2 = 0.21332 x 46.0 = 9.81272 ( 9.81 grms ) 8.48 g, 9.81 g

  14. 10. A blast furnace can produce about 700 tonnes of iron a day. • How much iron (III) oxide will be consumed? • Assuming coke is pure carbon, how much coke would be needed to produce the necessary carbon monoxide? • Fe2O3(s) + 3CO(g)  2Fe(l) + 3CO2(g) • 2C(s) + O2(g) 2CO(g) Moles Fe= 1000g = 1 Kg 1000 kg = 1 tonne so 1 tonne is 1.00 x106 grms 700 x 1000 x 1000 / 55.8 = 1.25448 x 107 ( 1.25 x 107 moles) Ratio Fe : Fe2O3 2 : 1 Moles : Fe2O3 = 1.2545 x 107 / 2 = 6.2724 x 106 ( 6.27 x 106 moles) Mass : Fe2O3= 159.6 x 6.2725 x 107 = 1.001x 109( 1000 tonnes) Ratio Fe : CO 2 : 3 Moles : CO = 1.25448 x 107 x 1.5 = 1.882 x 107 Ratio C : CO 2 : 2 moles :C=1.882x107 so the mass of C will be= 1.882 x 107 x 12.0 = 2.2584 x 108 g or 226 tonnes

  15. Answers to 1.1.3 -1.1.9 exercises • a) 0.10 • b) 0.078 • c) 5500 • d) 0.16 • e) 0.022 • a) 3.6 g • b) 14.9 g • c) 5.6 g • d) 39.9 kg • e) 6.8 g

  16. Answers to 1.1.3 -1.1.9 exercises • a) 28 (N2) • b) 40 (Ca) • c) 160 (Br2) • d) 28 (N2) • e) 249.6 (CuSO4.5H2O) • a) 1.51 x 1022 • b) 3.42 x 1022 • c) 1.45 x 1022 • d) 2.15 x 1024 • a) 11.7 g • b) 110 g • c) 8.07 mg

  17. Answers to 1.1.3 -1.1.9 exercises • a) 28 (N2) • b) 40 (Ca) • c) 160 (Br2) • d) 28 (N2) • e) 249.6 (CuSO4.5H2O) • a) 1.51 x 1022 • b) 3.42 x 1022 • c) 1.45 x 1022 • d) 2.15 x 1024 • a) 11.7 g • b) 110 g • c) 8.07 mg

  18. Answers to 1.1.3 -1.1.9 exercises • 1.06 g • 729 g • 43.3 g, 11.0 g • 8.48 g, 9.81 g • 1000 tonnes, 226 tonnes • a) 51.7% b) 17.0 % c) 87.2 % • a) 51.7 % • b) 67.4% • c) 52.2 % so (b) most efficient

  19. Answers to 1.1.3 -1.1.9 exercises • a) 0.10 b) 0.078 c) 5500 d) 0.16 e) 0.022 • a) 3.6 g b) 14.9 g c) 5.6 g d) 39.9 kg e) 6.8 g • a) 28 (N2) b) 40 (Ca) c) 160 (Br2) d) 28 (N2) e) 249.6 (CuSO4.5H2O) • a) 1.51 x 1022 b) 3.42 x 1022 c) 1.45 x 1022 d) 2.15 x 1024 • a) 11.7 g b) 110 g c) 8.07 mg • 1.06 g • 7. 729 g • 8. 43.3 g, 11.0 g • 8.48 g, 9.81 g • 10. 1000 tonnes, 527 tonnes • 11. a) 51.7% b) 17.0 % c) 87.2 % • 12. a) 51.7 % b) 67.4% c) 52.2 % so (b) most efficient

  20. 0.00131 x 40.1 = 0. 0525 or 5.25 x 10-2 0.00131 x 24.0 = 0. 0314 or 3.14 x 10-2 dm3 There are x2 OH- ions per Ca(OH)2 so the concentration is doubled ! 0.00131 x 2 = 0. 00262 or 2.62 x 10-3 mole and that’s in 250cm3 0.00262/0.250 = 0.0150 mols / dm3 1.50 x 10-2 M

  21. The molar mass of Barium is 137. this is approximately 3 time the molar mass of Calcium. So give the same mass of each the barium must contain less moles and hence will release less moles of hydrogen The rate of reaction of Barium would be greater than calcium – more vigorous / exothermic. This is due to the electrons in the outer shell being further away from the nucleus, and as there is greater shielding, there are more completed shell, the attraction will be less.

  22. . 00262/0.250 = 0.0150 mol dm3 1.50 x 10-2 M

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