STOICHIOMETRY II Section 12.2: Chemical Calculations.

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Stoichiometry II Section 12.2: Chemical Calculations

Transcript of STOICHIOMETRY II Section 12.2: Chemical Calculations.

Page 1: STOICHIOMETRY II Section 12.2: Chemical Calculations.

Stoichiometry IISection 12.2: Chemical Calculations

Page 2: STOICHIOMETRY II Section 12.2: Chemical Calculations.

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.

Page 4: STOICHIOMETRY II Section 12.2: Chemical Calculations.

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 HImol H2

=21

⇒ 1 × mol HI = 2 × mol H2 ⇒ mol HI = 2 × 0.75 mol = 1.5 mol= 1.5 mol

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Sample Problem 12.2How 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

=21

⇒ 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 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 CH4.

① nCH4 =mCH4

MCH4

=100 g

16 g/mol= 6.25 mol

6.25

_____ g

Page 9: STOICHIOMETRY II Section 12.2: Chemical Calculations.

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 16 g/mol 32 g/mol 44 g/mol 18 g/mol

m 100. g _____ g

n _____ mol

_____ mol②

The second step is to use mol ratio to find the number of mols of CO2.

⇒ 1×nCO2 = 1×nCH4

6.25

② =1

1

nCO2

nCH4 ⇒ nCO2 = 6.25 mol

6.25

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_____ mol②

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 16 g/mol 32 g/mol 44 g/mol 18 g/mol

m 100. g _____ g

n _____ mol

① ③

The third step is to find the mass of CO2.

= (6.25 mol)(44 g/mol)

6.25

③ mCO2 = nCO2 × MCO2 = 275 g

6.25

275

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Sample Problem 12.3Calculate 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 28 g/mol 2 g/mol 17 g/mol

m 5.40 g _______ g

_______ mol②

n _______ mol

①③

① nH2 =mH2

MH2

=5.40 g

2.00 g/mol= 2.70 mol

2.70

⇒ 3×nNH3 = 2×nH2 ② =

2

3

nNH3

nH2 ⇒ nNH3 = (2)(2.70 mol)/(3) = 1.80 mol

1.80

= (1.80 mol)(17 g/mol) ③ mNH3 = nNH3 × MNH3 = 30.6 g

30.6

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Additional Sample Problem 1Calculate 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 28 g/mol 2 g/mol 17 g/mol

m 745 g _______ g

_______ mol②

n _______ mol

①③

① nN2 =mN2

MN2

=745 g

28.0 g/mol= 26.6 mol

26.6

⇒ 1×nNH3 = 2×nN2 ② =

2

1

nNH3

nN2 ⇒ nNH3 = (2)(26.6 mol) = 53.2 mol

53.2

= (53.2 mol)(17.0 g/mol) ③ mNH3 = nNH3 × MNH3 = 905 g

905

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Additional Sample Problem 2Calculate 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 58.1 g/mol 32.0 g/mol 44.0 g/mol 18.0 g/mol

m 25.0 g ______ g

______ mol②

n ______ mol

① ③

① nC4H10 =mC4H10

MC4H10

=25.0 g

58.1 g/mol= 0.430 mol

0.430

⇒ 2×nO2 = 13×nC4H10 ② =

13

2

nO2

nC4H10 ⇒ nO2 = (13/2)(0.430 mol) = 2.80 mol

2.80

= (2.80 mol)(32.0 g/mol) ③ mO2 = nO2 × MO2 = 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).

mA nA =

mA

MA

mB = nBMBnA → nB

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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 12.5Nitrogen 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)

⇒ 1×VNO2 = 2×VO2 ① =

2

1

VNO2

VO2 ⇒ VNO2 = (2)(34 L) = 68 L

V 34 L ____ L①

68

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Sample Problem 12.6Assuming 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)

⇒ 2×VO2 = 1×VSO3 ① =

1

2

VO2

VSO3 ⇒ VO2 = (½)(20.4 mL) = 10.2 mL

V 20.4 mL____ mL①

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

Page 19: STOICHIOMETRY II Section 12.2: Chemical Calculations.

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.