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**Section 12.2: Chemical Calculations**

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 N2(g) + 3 H2(g) → 2 NH3(g), the mol ratio of N2 to H2 is 1 to 3 the mol ratio of N2 to NH3 is 1 to 2 the mol ratio of H2 to NH3 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 H2(g) + I2(g) → 2 HI(g), if we wanted to know the amount of HI produced from 0.75 mols of H2 with excess I2, we would use the mol ratio. mol HI mol H2 2 1 = ⇒ 1 × mol HI = 2 × mol H2 ⇒ mol HI = 2 × 0.75 mol = 1.5 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: N2(g) + 3 H2(g) → 2 NH3(g) mols N2 = nN2 = 0.60 mol Unknown: mols NH3 = nNH3 = ? Mol Solution: nNH3 nN2 2 1 = ⇒ 1 × nNH3 = 2 × nN2 ⇒ nNH3 = 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 CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l), what mass of CO2 is produced by the burning of 100. g of CH4? CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) M g/mol g/mol g/mol g/mol m g _____ g n _____ mol ① 6.25 Next, we write the known values and the unknown below the equation. The first step is to calculate the number of mols of CH4. First, we write the molar masses, M, above the symbols in the equation. ① nCH4 = mCH4 MCH4 = 100 g 16 g/mol = 6.25 mol

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**Writing and Using Mole Ratios**

For example, in the reaction CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l), what mass of CO2 is produced by the burning of 100. g of CH4? CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) M g/mol g/mol g/mol g/mol m g _____ g n _____ mol ① 6.25 _____ mol ② 6.25 The second step is to use mol ratio to find the number of mols of CO2. ② = 1 nCO2 nCH4 ⇒ 1×nCO2 = 1×nCH4 ⇒ nCO2 = 6.25 mol

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**Writing and Using Mole Ratios**

For example, in the reaction CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l), what mass of CO2 is produced by the burning of 100. g of CH4? CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) M g/mol g/mol g/mol g/mol m g _____ g 275 n _____ mol ① ③ 6.25 _____ mol ② 6.25 The third step is to find the mass of CO2. ③ mCO2 = nCO2 × MCO2 = (6.25 mol)(44 g/mol) = 275 g

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Sample Problem 12.3 Calculate the mass of NH3 produced by the reaction of 5.40 g of H2 with an excess of N2. The equation is: N2(g) + 3 H2(g) → 2 NH3(g). N2(g) + 3 H2(g) → 2 NH3(g) M g/mol 2 g/mol g/mol n _______ mol ① m g _______ g 30.6 ③ 2.70 _______ mol ② 1.80 ① nH2 = mH2 MH2 = 5.40 g 2.00 g/mol = 2.70 mol ② = 2 3 nNH3 nH2 ⇒ 3×nNH3 = 2×nH2 ⇒ nNH3 = (2)(2.70 mol)/(3) = 1.80 mol ③ mNH3 = nNH3 × MNH3 = (1.80 mol)(17 g/mol) = 30.6 g

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**Additional Sample Problem 1**

Calculate the mass of NH3 produced by the reaction of 745 g of N2 with an excess of H2. The equation is: N2(g) + 3 H2(g) → 2 NH3(g). N2(g) + 3 H2(g) → 2 NH3(g) M g/mol 2 g/mol g/mol n _______ mol ① m g _______ g 905 ③ 26.6 _______ mol ② 53.2 ① nN2 = mN2 MN2 = 745 g 28.0 g/mol = 26.6 mol ② = 2 1 nNH3 nN2 ⇒ 1×nNH3 = 2×nN2 ⇒ nNH3 = (2)(26.6 mol) = 53.2 mol ③ mNH3 = nNH3 × MNH3 = (53.2 mol)(17.0 g/mol) = 905 g

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**Additional Sample Problem 2**

Calculate the mass of O2 needed for the reaction of 25.0 g of C4H10. The equation is: 2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g). 2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g) M g/mol g/mol g/mol g/mol m g n ______ mol ① ______ g 89.5 ③ 0.430 ______ mol ② 2.80 ① nC4H10 = mC4H10 MC4H10 = 25.0 g 58.1 g/mol = mol ② = 13 2 nO2 nC4H10 ⇒ 2×nO2 = 13×nC4H10 ⇒ nO2 = (13/2)(0.430 mol) = 2.80 mol ③ mO2 = nO2 × MO2 = (2.80 mol)(32.0 g/mol) = 89.5 g

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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). mA nA = mA MA nA → nB mB = nBMB

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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 N2(g) + 3 H2(g) → 2 NH3(g), the volume ratio at STP is the same as the mol ratio. For every liter of N2(g) used, three liters of H2(g) would be used and two liters of NH3(g) would be produced. If we used two liters of N2(g), we would need six liters of H2(g) and we would produce four liters of NH3(g). If we used one-half liter of N2(g), we would need one and a half liters of H2(g) and we would produce one liter of NH3(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) + O2(g) → 2 NO2(g) V L ____ L ① 68 ① = 2 1 VNO2 VO2 ⇒ 1×VNO2 = 2×VO2 ⇒ VNO2 = (2)(34 L) = 68 L

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Sample Problem 12.6 Assuming STP, how many milliliters of oxygen are needed to produce 20.4 mL SO3 according to this balanced equation? 2 SO2(g) + O2(g) → 2 SO3(g) V mL ____ mL ① 10.2 ① = 1 2 VO2 VSO3 ⇒ 2×VO2 = 1×VSO3 ⇒ VO2 = (½)(20.4 mL) = 10.2 mL

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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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Stoichiometry Chapter 12.

Stoichiometry Chapter 12.

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