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Chapter03rw-final - 3 3.1 3.2 Stoichiometry Calculations...

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38 3 Stoichiometry: Calculations with Chemical Formulas and Equations Visualizing Concepts 3.1 Reactant A = red, reactant B = blue Overall, 4 red A 2 molecules + 4 blue B atoms 4 A 2 B molecules Since 4 is a common factor, this equation reduces to equation (a). 3.2 Write the balanced equation for the reaction. 2H 2 + CO CH 3 OH The combining ratio of H 2 : CO is 2:1. If we have 8 H 2 molecules, 4 CO molecules are required for complete reaction. Alternatively, you could examine the atom ratios in the formula of CH 3 OH, but the balanced equation is most direct. 3.3 (a) There are twice as many O atoms as N atoms, so the empirical formula of the original compound is NO 2 . (b) No, because we have no way of knowing whether the empirical and molecular formulas are the same. NO 2 represents the simplest ratio of atoms in a molecule but not the only possible molecular formula. 3.4 The box contains 4 C atoms and 16 H atoms, so the empirical formula of the hydro- carbon is CH 4 . 3.5 (a) Analyze . Given the molecular model, write the molecular formula. Plan . Use the colors of the atoms (spheres) in the model to determine the number of atoms of each element. Solve . Observe 2 gray C atoms, 5 white H atoms, 1 blue N atom, 2 red O atoms. C 2 H 5 NO 2
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3 Stoichiometry Solutions to Exercises 39 (b) Plan . Follow the method in Sample Exercise 3.9. Calculate formula weight in amu and molar mass in grams. 2 C atoms = 2(12.0 amu) = 24.0 amu 5 H atoms = 5(1.0 amu) = 5.0 amu 1 N atoms = 1(14.0 amu) = 14.0 amu 2 O atoms = 2(16.0 amu) = 32.0 amu 75.0 amu Formula weight = 75.0 amu, molar mass = 75.0 g/mol (c) Plan . The molar mass of a substance provides the factor for converting moles to grams (or grams to moles). Solve . 3 mol glycine × glycine g 225 mol glycine g 75.0 = (d) Plan . Use the definition of mass % and the results from parts (a) and (b) above to find mass % N in glycine. Solve . 100 NO H C g N g N % mass 2 5 2 × = Assume 1 mol C 2 H 5 NO 2 . From the molecular formula of glycine [part (a)], there is 1 mol N/mol glycine. % 7 . 18 100 g 75.0 g 0 . 14 100 glycine mass molar N) mass (molar 1 N % mass = × = × × = 3.6 Analyze . Given: 4.0 mol CH 4 . Find: mol CO and mol H 2 Plan . Examine the boxes to determine the CH 4 :CO mol ratio and CH 4 :H 2 O mole ratio. Solve . There are 2 CH 4 molecules in the reactant box and 2 CO molecules in the product box. The mole ratio is 2:2 or 1:1. Therefore, 4.0 mol CH 4 can produce 4.0 mol CO. There are 2 CH 4 molecules in the reactant box and 6 H 2 molecules in the product box. The mole ratio is 2:6 or 1:3. So, 4.0 mol CH 4 can produce 12:0 mol H 2 . Check . Use proportions. 2 mol CH 4 /2 mol CO = 4 mol CH 4 /4 mol CO; 2 mol CH 4 /6 mol H 2 = 4 mol CH 4 /12 mol H 2 . 3.7 Analyze . Given a box diagram and formulas of reactants, draw a box diagram of products. Plan . Write and balance the chemical equation. Determine combining ratios of elements and decide on limiting reactant. Draw a box diagram of products, containing the correct number of product molecules and only excess reactant.
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