ChemistryChemical EquationsBalancing Chemical EquationsWater balancingWater balancingWater balancingWater balancingRusting of IronProblem 1 in classN2 + H2 NH3Combustion of methaneProblem 2 in classC3H8 + O2 CO2 + H2OPoly-atomic GroupsPoly-atomic GroupsPoly-atomic GroupsVolume RelationshipsAvogadro’s hypothesisCombustion of propaneProblem 3 in classCombustion of propaneAtomic vs. Molecular WeightCalculate Molar WeightProblem 4 in classMolar mass of propane C3H8Moles calculated from GramsMoles calculated from GramsMoles calculated from GramsProblem 5 in classProblem 5 in classChemistryReading: Supplemental Text Materials pages 173-185Chemical Equations•C+O2 ÆCO2•C(s) +O2 (g) ÆCO2 (g)• Reactants on left, products on right• Each are balanced because same number of atoms of reactants as productsBalancing 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 sideWater balancingWater balancingWater balancingWater balancingRusting of Iron•Fe + O2Æ Fe2O3not balanced• start with oxygen• Fe + 3 O2Æ2 Fe2O3• next do iron•4 Fe + 3 O2Æ2 Fe2O3Problem 1 in classBalance the following equation, on the ½sheet worksheet provided. (Put your name on the back of the sheet, please.)N2 + H2Æ NH3N2 + H2Æ NH3•N2 + H2Æ NH3fix hydrogen•N2 + 3 H2Æ 2 NH3 • When H is fixed, N is already done!Combustion of methane•CH4+ O2Æ CO2 + H2O • fix hydrogen•CH4+ O2Æ CO2 + 2 H2O • need more O reactants•CH4+ 2 O2Æ CO2 + 2 H2O• Now check to see if you are doneProblem 2 in classBalance the following chemical equation, the combustion of propaneC3H8+ O2Æ CO2+ H2OC3H8+ O2Æ CO2+ H2OC3H8+ O2Æ CO2+ H2O • fix carbon•C3H8+ O2Æ 3 CO2+ H2O• Work on hydrogen•C3H8+ O2Æ 3 CO2+ 4 H2O• Now there are 10 oxygen on right•C3H8+ 5 O2Æ 3 CO2+ 4 H2OPoly-atomic Groups•SO4 is sulfate, common -2 ionic group•PO4 is Phosphate -3 ion• OH is Hydroxyl -1 ion• Combine with positive ions to balance charges to zero, as a group•Ca3(PO4)2(Ca is +2)Poly-atomic Groups•H2SO4+ NaCN Æ HCN + Na2SO4• Fix sodium first•H2SO4+ 2 NaCN Æ HCN + Na2SO4• Now Carbon and Nitrogen not balanced•H2SO4+ 2 NaCN Æ 2 HCN + Na2SO4• This also fixes the Hydrogen, SO4 is finePoly-atomic Groups•H3PO4+ Ca(OH)2Æ Ca3(PO4)2+ H2O•H3 (PO4) + Ca(OH)2Æ Ca3(PO4)2+ H2O•H3 (PO4) +3 Ca(OH)2Æ Ca3(PO4)2+ H2O•2H3 (PO4) +3 Ca(OH)2Æ Ca3(PO4)2+ H2O•2H3 (PO4) +3 Ca(OH)2Æ Ca3(PO4)2+6H2OVolume 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 ratiosAvogadro’s hypothesisCombustion of propaneC3H8+ 5 O2Æ 3 CO2+ 4 H2O• How much oxygen is needed to burn 0.556 L of propane?• Ratio of Oxygen molecules to propane molecules is 5:1• 0.556 L x 5 L O2= 2.87 L Oxygen1 L PropaneProblem 3 in class• Combustion of propaneC3H8+ 5 O2Æ 3 CO2+ 4 H2OCalculate how much CO2 is produced when 2 L of propane is burnedCombustion of propaneC3H8+ 5 O2Æ 3 CO2+ 4 H2O• Ratio of CO2to Propane•3:1• 2 L Propane x 3 L CO2=6 L CO21 L PropaneAtomic 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• Molar mass is the mass of one mole of molecules– One atomic mass number of grams– 6.0221367x1023moleculesCalculate Molar Weight• O atomic weight 15.9996• O-16 atomic mass 16 u• Molecular oxygen O2 atomic mass 32 u• Molecular O2molar mass 32 g/mole•CO2molar mass– C=12 g/mole, O2=32 g/mole–CO2=12+32=44 g/moleProblem 4 in class• Calculate the molar mass of propane C3H8• assume molar mass of C = 12 g/mole• assume molar mass of H = 1 g/moleMolar mass of propane C3H8• 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 C3H8Moles calculated from Grams• 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 gMoles calculated from Grams• Number of moles of 176 g of CO2• Molar mass of of CO2=44 g/mole• If you multiply, you get a mess– 176 g x 44 g/mole results in units of g2/mole– Alerts you that you made an errorMoles calculated from Grams• Number of moles of 176 g of CO2• Molar mass of of CO2=44 g/mole• Instead divide 176 g by 44 g/mole• Or look at it as176 g x (1 mole44 g)= 4 moles CO2Problem 5 in classHow many moles of Si-28 are there in 140 g of Si-28?Problem 5 in classHow many moles of Si-28 are there in 140 g of Si-28?• 140 g (1 mole28 g)= 5 moles
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