Chemistry. Chapter 17

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Transcription:

Chemistry Chapter 17

Chemical Equations C+O 2 CO 2 C (s) +O 2 (g) CO 2 (g) Reactants on left, products on right Each are balanced because same number of atoms of reactants as products

Balancing Chemical Equations Molecules of reactants and products shown Cannot change the molecule Can change how many of them Cannot add or delete reactants or products Balanced equal number of same atoms on each side

Water balancing

Rusting of Iron Fe + O 2 Fe 2 O 3 start with oxygen not balanced Fe + 3 O 2 2 Fe 2 O 3 next do iron 4 Fe + 3 O 2 2 Fe 2 O 3

Problem 1 in class Balance the following equation, on the worksheet provided. (Put your name on the back of the sheet, please.) N 2 + H 2 NH 3

Problem 1: N 2 + H 2 NH 3 N 2 + H 2 NH 3 fix hydrogen N 2 + 3 H 2 2 NH 3 When H is fixed, N is already done!

Combustion of propane C 3 H 8 + O 2 CO 2 + H 2 O fix hydrogen C 3 H 8 + O 2 CO 2 + 4 H 2 O need more carbon product C 3 H 8 + O 2 3 CO 2 + 4 H 2 O Do oxygen last, because it is single C 3 H 8 + 5 O 2 3 CO 2 + 4 H 2 O

Problem 2 in class Balance the following chemical equation, the combustion of methane CH 4 + O 2 CO 2 + H 2 O

Problem 2: CH 4 + O 2 CO 2 + H 2 O CH 4 + O 2 CO 2 + H 2 O fix hydrogen CH 4 + O 2 CO 2 +2 H 2 O Work on oxygen CH 4 + 2O 2 CO 2 + 2H 2 O Now check to see if you are done

Volume Relationships Equal volumes of gases at the same temperature and pressure have the same number of molecules Gases react in small whole number quantities Avogadro s hypothesis: chemicals react in consistent, small whole number ratios

Avogadro s hypothesis

Combustion of propane C 3 H 8 + 5 O 2 3 CO 2 + 4 H 2 O How much (what volume of) oxygen is needed to burn 0.5 L of propane? Ratio of Oxygen molecules to propane molecules is 5:1 0.5 L x 5 L O 2 = 2.5 L Oxygen 1 L Propane

Problem 3 in class Combustion of propane C 3 H 8 + 5 O 2 3 CO 2 + 4 H 2 O Calculate how much CO 2 is produced when 2 L of propane is burned

3: Combustion of propane C 3 H 8 + 5 O 2 3 CO 2 + 4 H 2 O Ratio of Propane to CO 2 1:3 2 L Propane x 3 L CO 2 = 6 L CO 2 1 L Propane

Atomic vs. Molecular Weight Atomic weight on periodic table is average of natural abundance of isotopes Atomic mass is the number of nucleons in a particular atom specified by isotope Molecular mass is the mass of one mole of molecules One atomic mass number of grams 6.0221367x10 23 molecules

Calculate Molecular Mass O atomic weight 15.9996 O-16 atomic mass 16 u Molecular oxygen O 2 atomic mass 32 u Molecular O 2 molecular mass 32 g/mole CO 2 molecular mass C=12 g/mole, O 2 =32 g/mole CO 2 =12+32=44 g/mole

Problem 4 in class Calculate the molar mass of propane C 3 H 8 assume molar mass of C = 12 g/mole assume molar mass of H = 1 g/mole

4: Molar mass of propane C 3 H 8 assume molar mass of C = 12 g/mole 3 C x 12 g/mole = 36 g/mole assume molar mass of H = 1 g/mole 8 H x 1 g/mole = 8 g/mole 8 + 36 = 44 g/mole C 3 H 8

Grams calculated from Moles Can find the molar mass of substance Na=23 g/mole Multiply molar mass times moles 23 g/mole x 0.250 moles = 5.75 g

Moles calculated from Grams 176 g of CO 2 = Number of moles? Molar mass of CO 2 = 44 g/mole If you multiply, you get a mess 176 g x 44 g/mole results in units of g 2 /mole UNITS Alert you that you made an error

Moles calculated from Grams 176 g of CO 2 = Number of moles? Molar mass of CO 2 = 44 g/mole Instead divide 176 g by 44 g/mole Same as multiply by reciprocal 176 g x ( 1 mole 44 g ) = 4 moles CO 2

Problem 5 in class 132 g of propane (C 3 H 8 )is how many moles? 44 g/mole propanec 3 H 8 1mole 132 g = 3moles 44g

Mass to moles relationship

Proportions Mathematical device to compare ratios Be sure you keep same:same in columns and rows Example moles of a g of a moles of b = g of b or moles of a = moles of b g of a g of b

Proportions correctly organized cross-multiply to solve

Problem NO + O NO 2 2 30 grams NO How many grams NO 2 produced? Balance Equation 2NO + O NO 2 2 2

Problem 2NO + O NO 2 2 2 30 grams NO How many grams NO 2 produced? Balance Equation Determine molar ratios 2:1:2 Find molar mass of each component NO=30 g, NO 2 =46 g, N=14 g, O=16 g,

Problem 2NO + O NO 2 2 2 30 grams NO How many grams NO 2 produced? Find molar mass of each component NO=30 g, NO 2 =46 g, N=14 g, O=16 g How many moles is 30 grams NO? One mole

Problem 2NO + O NO 2 2 2 30 grams NO How many grams NO 2 produced? Molar mass of each component N=14 g, O=16 g, NO=30 g, NO 2 =46 g One mole of NO Molar ratios 2:1:2 So one mole of NO 2 is produced How many grams is that?

Problem H S + O SO + H O 2 2 2 2 32 grams SO 2 How many grams O 2 used?

Problem H S + O SO + H O 2 2 2 2 Balance first 2 H S + 3O 2SO + 2H O 2 2 2 2 Then determine molar ratios 2 SO 2 to 3 O 2

Problem 2 H S + 3O 2SO + 2H O 2 2 2 2 Find molar masses (S)= 32+ (O 2 )= 32 = 64 g/mol SO 2 (H)= 2x1 + (O)= 16 = 18 g/mol H 2 O (H)= 2x1=2 + (S)= 32= 34 g/mol H 2 S (O)= 2x16= 32 g/mol O 2

Problem 2 H S + 3O 2SO + 2H O 2 2 2 2 32 g SO 2 needs how many grams O 2? How many moles is 32 g SO 2? 1mole 32 g SO = 0.5 moles 64 g 2 SO 2

Problem 2 H S + 3O 2SO + 2H O 2 2 2 2 32 g SO 2 needs how many grams O 2? How many moles O 2 is needed? 0.5 moles SO 2 in 2:3 ratio with O 2 1mole 32 g SO = 0.5 moles 64 g 2 SO 2

Problem 6 C + O CO 2 2 Molar ratio 1:1:1 4 grams oxygen Grams carbon consumed? Grams carbon dioxide produced?

Problem 6 C + O CO 2 2 Molar ratio 1:1:1 4 grams oxygen 1 mole O 2 = 32 g 1mole 4g O = 0.125 moles O 2 32g 2

Problem 6 C + O CO 2 2 Molar ratio 1:1:1 0.125 moles O 2 in 1:1 ratio 0.125 moles C 0.125 moles CO 2

Problem 6 C + O CO 2 2 0.125 moles C Grams carbon consumed? 12g 0.125moles C = 1mole 1.5g C 44g 0.125 mole CO = 5.5g CO 2 1mole 2

Reaction Speed Collision of molecules required for it to occur Increase concentration Increase temperature Catalyst can facilitate reaction

Chlorine catalyst

Energy of reactions Release energy EXOTHERMIC Methane combustion Consume energy ENDOTHERMIC Formation of water