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Combustion analysis

Combustion analysis. 1 mol CO2 44 g CO2. 1 mol C atoms 1 mol CO2. = 0.10 mol C atom. 4.40 g CO2. 1 mol H2O 18 g H2O. 2 mol H atoms 1 mol H2O. = 0.20 mol H atom. 1.80 g H2O. 3.00g = weight O + weight C + weight H. weight O = 3.00g - weight C weight H.

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Combustion analysis

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  1. Combustion analysis 1 mol CO2 44 g CO2 1 mol C atoms 1 mol CO2 = 0.10 mol C atom 4.40 g CO2 1 mol H2O 18 g H2O 2 mol H atoms 1 mol H2O = 0.20 mol H atom 1.80 g H2O 3.00g = weight O + weight C + weight H weight O = 3.00g - weight C weight H weight O = 3.00g – 12g/mol(0.1molC) – 1g/mol(0.2molH) 1 mol O atom 16 g O = 1.60 g O = 0.10 mol O atom

  2. Combustion analysis (cont) Empirical formula: CH2O Original molecular weight: 180 g/mol Formula weight of CH2O: 30 g/mol 6 CH2O units required to get to formula weight. Thus the molecular formula is: C6H12O6

  3. Acetic Acid and Ethanol 1 mol acetic 60 g acetic 1 mol ester 1 mol acetic 88g ester 1 mol ester = 6.16g ester 4.20 g Acetic Total of 6.16 g of ethyl acetate can be made from this reaction

  4. Iron(III) Oxide + Aluminum 1 mol Fe2O3 159.7 g Fe2O3 2 mol Fe 1 mol Fe2O3 55.84g Fe 1 mol Fe = 106 g Fe 152 g Fe2O3 1 mol Al 26.98 g Al 2 mol Fe 2 mol Al 55.84g Fe 1 mol Fe 74.2 g Al = 154 g Fe Thus the Fe2O3 is the limiting reagent

  5. Iron(III) Oxide + Aluminum 1 mol Fe2O3 159.7 g Fe2O3 2 mol Al 1 mol Fe2O3 26.98g Al 1 mol Al = 51.4 g Al 152 g Fe2O3 Aluminum Remaining = Starting – Used Remaining = 74.2 g Al – 51.4 g Al = 22.8 g Al

  6. Titanium(IV) Chloride 1 mol TiO2 79.9 g TiO2 3 mol TiCl4 3 mol TiO2 189.7gTiCl4 1 mol TiCl4 = 9.82 g TiCl4 4.15 g TiO2 189.7gTiCl4 1 mol TiCl4 1 mol C 12.01 g C 3 mol TiCl4 4 mol C 5.67 g C = 67.2 g TiCl4 1 mol Cl2 70.9 g Cl2 3 mol TiCl4 6 mol Cl2 189.7gTiCl4 1 mol TiCl4 6.78 g Cl2 = 9.07 g TiCl4 Thus the Cl2 is the limiting reagent.

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