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STOICHIOMETRY II Section 12.2: Chemical Calculations

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Objectives Upon completion of this presentation, you will be able to construct mole ratios from balanced equations. apply mole ratios in stoichiometric calculations. calculate stoichiometric quantities from balanced chemical equations using units of moles, mass, representative particles, and volumes at STP.

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Writing and Using Mole Ratios A balanced chemical equation contains a lot of quantitative information. We can relate particles, moles, and masses. The most important thing that a properly balanced equation gives us is the number of moles for each of the reactants and products. From this, we get the mol ratio. The mol ratio is a conversion factor derived from the coefficients of the balanced equation.

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Writing and Using Mole Ratios The mol ratio is a conversion factor derived from the coefficients of the balanced equation. For example, in the equation N 2 (g) + 3 H 2 (g) 2 NH 3 (g), the mol ratio of N 2 to H 2 is 1 to 3 the mol ratio of N 2 to NH 3 is 1 to 2 the mol ratio of H 2 to NH 3 is 3 to 2

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Writing and Using Mole Ratios In chemical calculations, the mol ratios are used to convert between mols of reactant and mols of product mols of reactants mols of products For example, in the reaction H 2 (g) + I 2 (g) 2 HI(g), if we wanted to know the amount of HI produced from 0.75 mols of H 2 with excess I 2, we would use the mol ratio. mol HI mol H 2 = × mol HI = 2 × mol H 2 mol HI = 2 × 0.75 mol= 1.5 mol

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Sample Problem 12.2 How many mols of ammonia are produced when 0.60 mols of nitrogen react with hydrogen? Known:N 2 (g) + 3 H 2 (g) 2 NH 3 (g) mols N 2 = n N2 = 0.60 mol Unknown:mols NH 3 = n NH3 = ? Mol Solution: n NH3 n N2 = × n NH3 = 2 × n N2 n NH3 = 2 × 0.60 mol= 1.2 mol

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Writing and Using Mole Ratios We can also use mol ratios to convert between mass of reactant and mass of product masses of reactants masses of products

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Writing and Using Mole Ratios For example, in the reaction CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(l), what mass of CO 2 is produced by the burning of 100. g of CH 4 ? CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(l) M 16 g/mol 32 g/mol 44 g/mol 18 g/mol m 100. g n _____ mol First, we write the molar masses, M, above the symbols in the equation.Next, we write the known values and the unknown below the equation. The first step is to calculate the number of mols of CH 4. n CH4 = m CH4 M CH4 = 100 g 16 g/mol = 6.25 mol 6.25 _____ g

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Writing and Using Mole Ratios For example, in the reaction CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(l), what mass of CO 2 is produced by the burning of 100. g of CH 4 ? CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(l) M 16 g/mol 32 g/mol 44 g/mol 18 g/mol m 100. g_____ g n _____ mol _____ mol The second step is to use mol ratio to find the number of mols of CO 2. 1×n CO2 = 1×n CH = 1 1 n CO2 n CH4 n CO2 = 6.25 mol 6.25

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_____ mol Writing and Using Mole Ratios For example, in the reaction CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(l), what mass of CO 2 is produced by the burning of 100. g of CH 4 ? CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(l) M 16 g/mol 32 g/mol 44 g/mol 18 g/mol m 100. g_____ g n _____ mol The third step is to find the mass of CO 2. = (6.25 mol)(44 g/mol) 6.25 m CO2 = n CO2 × M CO2 = 275 g

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Sample Problem 12.3 Calculate the mass of NH 3 produced by the reaction of 5.40 g of H 2 with an excess of N 2. The equation is: N 2 (g) + 3 H 2 (g) 2 NH 3 (g). N 2 (g) + 3 H 2 (g) 2 NH 3 (g) M 28 g/mol 2 g/mol 17 g/mol m 5.40 g _______ g _______ mol n _______ mol n H2 = m H2 M H2 = 5.40 g 2.00 g/mol = 2.70 mol ×n NH3 = 2×n H2 = 2 3 n NH3 n H2 n NH3 = (2)(2.70 mol)/(3) = 1.80 mol 1.80 = (1.80 mol)(17 g/mol) m NH3 = n NH3 × M NH3 = 30.6 g 30.6

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Additional Sample Problem 1 Calculate the mass of NH 3 produced by the reaction of 745 g of N 2 with an excess of H 2. The equation is: N 2 (g) + 3 H 2 (g) 2 NH 3 (g). N 2 (g) + 3 H 2 (g) 2 NH 3 (g) M 28 g/mol 2 g/mol 17 g/mol m 745 g _______ g _______ mol n _______ mol n N2 = m N2 M N2 = 745 g 28.0 g/mol = 26.6 mol ×n NH3 = 2×n N2 = 2 1 n NH3 n N2 n NH3 = (2)(26.6 mol) = 53.2 mol 53.2 = (53.2 mol)(17.0 g/mol) m NH3 = n NH3 × M NH3 = 905 g 905

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Additional Sample Problem 2 Calculate the mass of O 2 needed for the reaction of 25.0 g of C 4 H 10. The equation is: 2 C 4 H 10 (g) + 13 O 2 (g) 8 CO 2 (g) + 10 H 2 O(g). 2 C 4 H 10 (g) + 13 O 2 (g) 8 CO 2 (g) + 10 H 2 O(g) M 58.1 g/mol 32.0 g/mol 44.0 g/mol 18.0 g/mol m 25.0 g ______ g ______ mol n ______ mol n C4H10 = m C4H10 M C4H10 = 25.0 g 58.1 g/mol = mol ×n O2 = 13×n C4H10 = 13 2 n O2 n C4H10 n O2 = ( 13 / 2 )(0.430 mol) = 2.80 mol 2.80 = (2.80 mol)(32.0 g/mol) m O2 = n O2 × M O2 = 89.5 g 89.5

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Other Stoichiometric Calculations In a typical stoichiometric problem, a given quantity (almost always mass) is first converted to mols. Then the mol ratio from the balanced equation is used to calculate the number of mols of the desired substance. Finally, the number of mols of the desired substance is converted to the proper quantity (usually mass). mAmA n A = mAmA MAMA m B = n B M B n A n B

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Other Stoichiometric Calculations In volume to volume stoichiometric calculations, there is no need to convert the volumes at STP to mols. The mol ratio is used to directly convert from the volume of one substance to the volume of another substance. For example: in the reaction N 2 (g) + 3 H 2 (g) 2 NH 3 (g), the volume ratio at STP is the same as the mol ratio. For every liter of N 2 (g) used, three liters of H 2 (g) would be used and two liters of NH 3 (g) would be produced. If we used two liters of N 2 (g), we would need six liters of H 2 (g) and we would produce four liters of NH 3 (g). If we used one-half liter of N 2 (g), we would need one and a half liters of H 2 (g) and we would produce one liter of NH 3 (g).

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Sample Problem 12.5 Nitrogen monoxide and oxygen gas combine to form the brown gas nitrogen dioxide, which contributes to photochemical smog. How many liters of nitrogen dioxide are produced when 34 liters of oxygen reacts with excess nitrogen monoxide? Assume conditions of STP. 2 NO(g) + O 2 (g) 2 NO 2 (g) 1×V NO2 = 2×V O2 = 2 1 V NO2 V O2 V NO2 = (2)(34 L) = 68 L V 34 L ____ L 68

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Sample Problem 12.6 Assuming STP, how many milliliters of oxygen are needed to produce 20.4 mL SO 3 according to this balanced equation? 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) 2×V O2 = 1×V SO3 = 1 2 V O2 V SO3 V O2 = (½)(20.4 mL) = 10.2 mL V 20.4 mL ____ mL 10. 2

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Summary The mol ratio is a conversion factor derived from the coefficients of the balanced equation. In chemical calculations, the mol ratios are used to convert between mols of reactant and mols of product mols of reactants mols of products We can also use mol ratios to convert between mass of reactant and mass of product masses of reactants masses of products

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Summary In a typical stoichiometric problem, a given quantity (almost always mass) is first converted to mols. Then the mol ratio from the balanced equation is used to calculate the number of mols of the desired substance. Finally, the number of mols of the desired substance is converted to the proper quantity (usually mass). In volume to volume stoichiometric calculations, there is no need to convert the volumes at STP to mols. The mol ratio is used to directly convert from the volume of one substance to the volume of another substance.

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