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3 Stoichiometry Solutions to Exercises (c) 3.107 (a) Analyze. Given: gasoline = C₈H₁₈ density = 0.69 g/mL, 20.5 mi/gal, 225 mi. Find: kg Plan. Write and balance the equation for the combustion of octane. Change mi gal octane mL g octane. Use stoichiometry to calculate g and kg from g octane. Solve. 225 mi 20.5 mi 1 gal X 1 mL 0.69 1 mL octane = = 29 kg octane 2.8667 C₈H₁₈ 114.2 1 mol C₈H₁₈ C₈H₁₈ 2 16 mol mol C₈H₁₈ 44.01 1 mol = 8.8382 10⁴ g = Check. 20 10³ = (45 0.7) = = kg octane ≈ (b) Plan. Use the same strategy as part (a). Solve. 225 mi 5 mi 3.7854 1 gal L 1 mL 0.69 1mL octane = 1.1754 10⁵ = kg octane 1.1754 X C₈H₁₈ 114.2 1 mol g C₈H₁₈ C₈H₁₈ 2 16 mol mol C₈H₁₈ 1 mol CO₂ = Check. Mileage of 5 mi/gal requires ~4 times as much gasoline as mileage of 20.5 mi/gal, so it should produce ~4 times as much 90 kg [from (a)] 4 = 360 = 4 10² kg [from (b)]. 3.108 Structural isomers, like 1-propanol and 2-propanol, have the sa me number and kinds of atoms, but different arrangements of these atoms. Since molecular weight is the sum of atomic weights, and number and kinds of atoms are the same, the molecular weights of structural isomers are the same. Again, because number and kinds of atoms are the same, percent composition and therefore combustion analysis results will be the same. Physical properties, like boiling point and density, are influenced by structure as well as molecular weight, and are different for structural isomers. The properties (a) boiling point and (d) density will distinguish between 1-propanol and 2-propanol. This is confirmed by comparing these properties from either Wolfram Alpha (WA) or the CRC Handbook of Chemistry and Physics (CRC). 75