9. 2 c 3 h 7 oh + 9 o 2 → 6 co 2 + 8 h 2 o

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2 C 3 H 7 OH + 9 O 2 6 CO 2 + a) 6 mol CO 2 2 mol C 3 H 7 OH c) 9 mol O 2 6 mol CO 2 b) 2 mol C 3 H 7 OH 9 mol O 2 d) 8 mol H 2 O 2 mol C 3 H 7 OH e) 6 mol CO 2 8 mol H 2 O f) 8 mol H 2 O 9 mol O 2 tify some mole ratios, using the given equation for ustion of isopropyl alchol.

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9. 2 C 3 H 7 OH + 9 O 2 → 6 CO 2 + 8 H 2 O. Identify some mole ratios, using the given equation for the combustion of isopropyl alchol. a) 6 mol CO 2 2 mol C 3 H 7 OH. d) 8 mol H 2 O 2 mol C 3 H 7 OH. b ) 2 mol C 3 H 7 OH 9 mol O 2. e ) 6 mol CO 2 - PowerPoint PPT Presentation

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9. 2 C3H7OH + 9 O2 6 CO2 + 8 H2Oa)6 mol CO2 2 mol C3H7OHc)9 mol O2 6 mol CO2

b)2 mol C3H7OH9 mol O2

d)8 mol H2O 2 mol C3H7OH

e)6 mol CO2 8 mol H2O

f)8 mol H2O 9 mol O2

Identify some mole ratios, using the given equation for the combustion of isopropyl alchol. Mole Ratios

3 CaCl2 + 2 H3PO4 Ca3(PO4)2 + 6 HCl

Identify some mole ratios, using the given equation. a)3 CaCl2 b) 6 HClc)3 CaCl2 1 Ca3(PO4)2 2 H3PO4 2 H3PO4 d)1 Ca3(PO4)2 e) 6 HClf)2 H3PO4 2 H3PO4 1 Ca3(PO4)2 6 HCl

11. C2H5OH + 3 O22 CO2 + 3 H2OHow many moles of CO2 can be produced from 7.75 mol C2H5OH?2 mol CO2 = X = 15.5 mol CO2 1 mol C2H5OH 7.75 mol C2H5OH

7.75 mol C2H5OH 2 mol CO2 = 15.5 mol CO2 1 mol C2H5OH

or4 HCl + O22 Cl2 + 2 H2OHow many moles of Cl2 can be produced from 5.60 mol HCl?2 mol Cl2 =X = 2.80 mol Cl2 4 mol HCl 5.60 mol HCl 5.60 mol HCl 2 mol Cl2 =2.80 mol Cl2 4 mol HClor13. MnO2 + 4 HCl Cl2 + MnCl2 + 2 H2OHow many moles of HCl will react with1.05 mol MnO2?1.05 mol MnO2 = 1 mol MnO2 X mol HCl 4 mol HCl

X = 4.20 mol HCl1.05 mol MnO2 4 mol HCl = 4.20 mol HCl 1 mol MnO2or14. Al4C3 + 12 H2O 4 Al(OH)3 + 3 CH4

How many moles of water are needed to react with 100. g Al4C3 ?

How many moles of Al(OH)3 will be produced when 0.600 mol of CH4 is formed?

100.0 g Al4C3 1 mol Al4C3 12 mol H2O = 8.333 mol H2O 144.0 g Al4C3 1 mol Al4C30.600 mol CH4 4 mol Al(OH)3 = 0.800 mol Al(OH)3 3 mol CH4

15. Ca(OH)2 + Na2CO3 2 NaOH + CaCO3 How many grams of sodium hydroxide can be produced from 500 g of calcium hydroxide according to this equation?1 mole Ca(OH) 2 = 74.1 g Ca(OH) 2 = 500 g 2 moles NaOH 80.0 g NaOH x74.1 x = 40000 x = 540 g NaOH

How many grams of zinc phosphate, Zn3(PO4)2, are formed when 10.0 g of Zn are reacted with phosphoric acid?3 Zn + 2 H3PO4 Zn3(PO4)2 + 3 H2 10.0 g? g3 mol Zn 1 mol Zn3(PO4)2= 196.2 g Zn 386.1 g Zn3(PO4)2196.2 g Zn = 10.0 g Zn 386.1 g Zn3(PO4)2 X g Zn3(PO4)2X = 19.7 g Zn3(PO4)217. Fe2O3 + 3C 2Fe + CO

How many kilograms of iron would be formed from 125 kg of Fe2O3?1 mole Fe2O3 = 0.1598 kg = 125 kg 2 mole Fe 0.1118 kg x0.1598 x = 13.975

x = 13.975 = 87.5 kg Fe 0.1598

18.How many grams of steam and iron must react to produce 375 g of magnetic iron oxide, Fe3O4?3 mol Fe 1 mol Fe3O43 Fe + 4 H2O Fe3O4 + 4 H2 4 mol H2O 1 mol Fe3O4 =167.4 g Fe 231.6 g Fe3O4 =72.0 g H2O231.6 g Fe3O4 =X g Fe 375 g Fe3O4 =X g H2O375 g Fe3O4X = 271 g Fe X = 117 g H2O19. a. 2C2H6 + 7O2 4CO2 + 6H2O

How many moles of O2 are needed for the complete combustion of 15.0 mol of ethane?7 mole O2 = X mole O22 mole C2H6 15 mole C2H62x = 105 x = 52.5 mole O2

19 b. 2C2H6 + 7O2 4CO2 + 6H2O

How many grams of CO2 are produced for each 8.00 g of H2O produced?

4 mole CO2 = 176 g CO2 = Xg CO26 mole H2O 108 g H2O 8.00gH2O108x = 1408

x = 1408 = 13.0 g CO2 108

19 c. 2C2H6 + 7O2 4CO2 + 6H2O

How many grams of CO2 will be produced by the combustion of 75.0 g of C2H6?

4 mole CO2 = 2 = 88.0 g CO2 = x 2 mole C2H6 1 30.0g C2H6 75.030x = 6600

x = 6600 = 220. g CO2 30

20.4 FeS2 + 11 O2 2 Fe2O3 + 8 SO2 How many moles of Fe2O3 can be made from 1.00 mol of FeS2 ?4 mol FeS2 2 mol Fe2O3 = 1.00 mol FeS2 X mol Fe2O3X = 0.500 mol Fe2O320.4 FeS2 + 11 O2 2 Fe2O3 + 8 SO2 (b) How many moles of O2 are required to react with 4.50 mol of FeS2 ?4 mol FeS2 11 mol O2 =4.50 mol FeS2 X mol O2X = 12.4 mol O220.4 FeS2 + 11 O2 2 Fe2O3 + 8 SO2 (c) If the reaction produces 1.55 mol of Fe2O3 , how many moles of SO2 are produced?2 mol Fe2O3 8 mol SO2 = 1.55 mol Fe2O3 X mol SO2 X = 6.20 mol SO2 20.4 FeS2 + 11 O2 2 Fe2O3 + 8 SO2 (d) How many grams of SO2 can be formed from 0.512 mol of FeS2 ?8 mol SO2 4 mol FeS2 0.512 mol FeS2 = 1.02 mol SO2 1.02 mol SO2 64.1 g SO2 1 mol SO2 = 65.4 g SO2 20.4 FeS2 + 11 O2 2 Fe2O3 + 8 SO2 (e) If the reaction produces 40.6 g of SO2 , how many moles of O2 were reacted?40.6 g SO2 11 mol O2 8 mol SO2 1 mol SO2 64.1 g SO2 = 0.633 mol SO2 0.633 mol SO2 = 0.870 mol O2 20.4 FeS2 + 11 O2 2 Fe2O3 + 8 SO2 (f) How many grams of FeS2 are needed to produce221 g of Fe2O3 ?221 g Fe2O31 mol Fe2O3159.7 g Fe2O3= 1.38 mol Fe2O31.38 mol Fe2O34 mol FeS2 2 mol Fe2O3120.0 g FeS2 1 mol FeS2 = 2.76 mol FeS2 2.76 mol FeS2 = 331 g FeS2 20.4 FeS2 + 11 O2 2 Fe2O3 + 8 SO2 (f) How many grams of FeS2 are needed to produce221 g of Fe2O3 ?4 mol FeS2 2 mol Fe2O3 =480.0 g FeS2 319.4 g Fe2O3 = X g FeS2 221 g Fe2O3X = 332 g FeS2 22. a)2 Bi(NO3)3 + 3 H2S Bi2S3 + 6 HNO350.0 g 6.00 g ? g2 Bi(NO3)3 = 790 g Bi(NO3)3 = 50.0 g Bi(NO3)3 1 Bi2S3 514.3 g Bi2S3 X grams Bi2S3

X = 32.6 grams

3 H2S = 102.3 g H2S = 6.00 g H2S 1 Bi2S3 514.3 g Bi2S3 X g Bi2S3

H2S limits X = 30.2 grams 22. b.) 3 Fe + 4 H2O Fe3O4 + 4 H2 40.0 g 16.0 g ? grams3Fe = 167.4 g Fe = 40.0 g Fe x= 1.91 g H24H2 8.0 g x g H2

4H2O = 1H2O = 18.0g H2O = 16.0 g H2O4H2 1H2 2.0 g H2 x g H2

H2O limiting x= 1.78 g H2

23) a)C3H8 + 5 O2 3 CO2 + 4 H2O 20.0 g 20.0 g ? mol20.0 g C3H8 1 mol C3H8 3 mol CO2 = 1.36 mol CO2 44.0 g C3H8 1 mol C3H8 20.0 g O2 1 mol O2 3 mol CO2 = 0.375 mol CO2 32.0 g O2 5 mol O2

20.0 g C3H8 1 mol C3H8 3 mol CO2 = 1.36 mol CO2 44.0 g C3H8 1 mol C3H8 80.0 g O2 1 mol O2 3 mol CO2 = 1.50 mol CO2 32.0 g O2 5 mol O2

23 b) C3H8 + 5 O2 3 CO2 + 4 H2O 20.0 g 80.0 g ? mol24.C3H8 + 5O2 3CO2 + 4H2O 5.0 mole 5.0 mole ? Moles1C3H8 = 5C3H8 X = 15 mole CO2 3 CO2 X

5O2 = 5 X = 3.0 mole CO23CO2 X O2 limits 24. C3H8 + 5O2 3CO2 + 4H2O3.0 mole 20. mole ? molesb) 1 mole C3H8 = 3.0 C3H83 mole CO2 XX = 9.00 mole CO2 C3H8 limits5 mole O2 = 20. O23 mole CO2 XX = 12 mole CO224. C3H8 + 5O2 3CO2 + 4H2O3.0 mole 20. mole ? moles24c) 1 C3H8 = 20 C3H83 CO2 X X = 60 mol CO25 O2 = 3 O23 CO2 XX = 1.8 mol CO2 O2 limits27) 2 Al + 3 Br2 2 AlBr3 25.0 g 100.0 g ? gIf 64.2 g of AlBr3 is recovered, what is the percent yield?2 mol Al = 54.0 g Al = 25.0 g Al X = 247 g2 mol AlBr3 533.4 g AlBr3 X g AlBr3 3 mol Br2 = 479.4 g Br2 = 100.0 g Br2 X = 111.3 g 2 mole AlBr3 533.4 g AlBr3 X g AlBr3% yield = Actual (lab) x 100 = 64.2 g = 57.7 % Calculated (theory) 111.3 g