ch1410 lecture #13 troch8 stoichiometry web...

Stoichiometry

The study of the numerical relationship between quantities in a chemical reaction

Making Pizza

The number of pizzas you can make depends on the amount of the ingredients you use.

1 crust + 5 oz. tomato sauce + 2 cu cheese ➜ 1 pizza

This relationship can be expressed mathematically 1 crust : 5 oz. sauce : 2 cu cheese : 1 pizza

If you want to make more than one pizza, you can use the amount of cheese you have to determine the number of pizzas you can make.

Predicting Amounts from Stoichiometry

The amounts of any other substance produced or consumed in a chemical reaction can be

determined from the amount of just one substance.

Moles of A

Moles of B

Grams of A

Grams of B

Particles of A

Particles of B

Avogadro’s Number Avogadro’s Number

Molar MassMolar Mass

Coefficients

1 mol glucose 6 mol water

6 mol water 1 mol glucose conversion factors

glucose + oxygen gas ➜ carbon dioxide + water

C6H12O6 + 6 O2 ➜ 6 CO2 + 6 H2O

1 mol C6H12O6 6 mol H2O

6 mol H2O 1 mol C6H12O6 conversion factors

1. According to the following equation, how many moles of water are made in the combustion of 0.10 moles of glucose?

0.10 mol C6H12O6 x = mol H2O6 mol H2O1 mol C6H12O6

0.60

mol C6H12O6 mol H2O

Moles of A

Moles of B

Mole to Mole Ratio from balanced equation

Stoichiometry Road Map

Grams of A

Grams of B

Molar Mass

Particles of B

Avogadro’s Number

Particles of A

3.5 x 1015 g C8H18 x x x

= 1.0789 x 1016 g CO2

16 mol CO22 mol C8H18

44.01 g CO21 mol CO2

1 mol C8H18114.22 g C8H18

2. Estimate the mass of CO2 produced in 2007 by the combustion of 3.5 x 1015 g of octane (C8H18).

2 C8H18(l) + 25 O2(g) ➜ 16 CO2(g) + 18 H2O(g)

2 mol C8H18 16 mol CO2

16 mol CO2 2 mol C8H18 conversion factors

1 mol C8H18 114.22 g C8H18

114.22 g C8H18 1 mol C8H18 conversion factors

1 mol CO2 44.01 g CO2

44.01 g CO2 1 mol CO2 conversion factors

1.1 x 1016

g C8H18 mol C8H18 mol CO2 g CO2

1 mol C6H12O6 6 mol CO2

6 mol CO2 1 mol C6H12O6

conversion factors

3. How many grams of glucose can be synthesized from 37.8 g of CO2 in photosynthesis?

6 CO2 + 6 H2O ➜ C6H12O6 + 6 O2

37.8 g CO2 x x x

= 25.796 g C6H12O6

1 mol CO244.01 g CO2

1 mol C6H12O66 mol CO2

180.2 g C6H12O61 mol C6H12O6

1 mol CO2 44.01 g CO2

44.01 g CO2 1 mol CO2 conversion factors

1 mol C6H12O6 180.2 g C6H12O6

180.2 g C6H12O6 1 mol C6H12O6 conversion factors

25.8

g CO2 mol CO2 mol C6H12O6 g C6H12O6

32.00 g O21.000 mol O2

1.000 mol O22.000 mol PbO2

1.000 mol PbO2239.2 g PbO2

100.0 g PbO2 x x x

= 6.689 g O2

2 PbO2(s) → 2 PbO(s) + O2(g)(PbO2 = 239.2, O2 = 32.00)

4. How many grams of O2 can be made from the decomposition of 100.0 g of PbO2?

1 mol PbO2 239.2 g PbO2

239.2 g PbO2 1 mol PbO2 conversion factors

2 mol PbO2 1 mol O2

1 mol O2 2 mol PbO2

conversion factors

1 mol O2 32.oo g O2

32.00 g O2 1 mol O2 conversion factors

g PbO2 mol PbO2 mol O2 g O2

Moles of A

Moles of B

Mole to Mole Ratio from balanced equation


Grams of A

Grams of B

Molar Mass

Particles of B

Avogadro’s Number

Particles of A

More Making Pizzas1 crust + 5 oz. tomato sauce + 2 cu cheese ➜1 pizza

What would happen if we had 4 crusts, 15 oz. tomato sauce, and 10 cu cheese?

Limiting reagent Theoretical

yield

The Limiting Reactant

For reactions with multiple reactants, it is likely that one of the reactants will be completely used before the others.

When this reactant is used up, the reaction stops and no more product is made.

Limiting ReactantThe reactant that is completely consumed in a chemical reaction

Theoretical YieldThe amount of product that can be made in a chemical reaction based on the

amount of limiting reactant

Actual YieldThe amount of product actually produced by a chemical reaction

Actual YieldPercent Yield = --------------------------------- x 100%

Theoretical Yield


Mole to Mole Ratio from balanced equationMoles of

CMole to Mole

Ratio from balanced

equation

Molar Mass

Moles of B

Moles of A

Grams of B

Grams of A

Molar Mass

Molar Mass

Grams of C

Compare the results of A and B

Use the smaller result to get a

theoretical yield

A + B C

Limiting and Excess Reactants in the Combustion of Methane

CH4(g) + O2(g) ➜ CO2(g) + H2O(g)

CH4(g) + 2 O2(g) ➜ CO2(g) + 2 H2O(g)

➜

8 molecules O2 x = 4 molecules of CO21 molecule CO22 molecules O2

5 molecules CH4 x = 5 molecules of CO21 molecule CO21 molecule CH4

➜

If we have five molecules of CH4 and eight molecules of O2, which is the limiting reactant?

8 mol O2 x = 4 mol of CO21 mol CO22 mol O2

➜

What happens when you mix five molecules of CH4 and eight molecules of O2?

1.00 mol Si3N4 2.00 mol N2

1.00 mol N2 x = 0.500 mol Si3N4

1.20 mol Si x = 0.400 mol Si3N41.00 mol Si3N4 3.00 mol Si

5. How many moles of Si3N4 can be made from 1.20 moles of Si and 1.00 mole of N2 in the reaction:

3 Si + 2 N2 ➜ Si3N4 ?

Theoretical yield

Limitingreactant

6. How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of copper(II) oxide?

2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)

smaller amount is from limiting reactant}

molNH3

g

mol

molN2

mol

mol

gNH3

gCuO

molCuO

molN2

45.2 g CuO x x = 0.1894 mol N2

Limiting reactant Theoretical yield

9.05 g NH3 x x = 0.2657 mol N2 1.00 mol NH317.03 g NH3

1.00 mol N22.00 mol NH3

1.00 mol CuO79.55 g CuO

1.00 mol N23.00 mol CuO



0.189 mol N2 x = 5.30 g N228.02 g N2

1.00 mol N2

More Making Pizzas

We can calculate the efficiency of making pizzas by calculating the percentage of the maximum number of

pizzas we actually make. In chemical reactions, we call this the percent yield.

Let’s now assume that as we are making pizzas, we burn a pizza, drop one on the floor, or other

uncontrollable events happen so that we only make two pizzas. This is the actual yield.

Actual YieldTheoretical Yield

x 100 % = Percent Yield

2 pizzas3 pizzas

x 100 % = 67%


2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l) If 4.61 g of N2 are made, what is the percent yield?

Theoretical Yield

smaller molN2

gN2

Theoretical Yield

Actual Yield

= % Yield

0.189 mol N2 x = 5.30 g N228.02 g N2

1.00 mol N2



If 4.61 g of N2 are isolated, what is the percent yield?

45.2 g CuO x x = 0.1894 mol N21.00 mol CuO79.55 g CuO

1.00 mol N23.00 mol CuO

Theoretical yield

percent yield

4.61 g N25.30 g N2

x 100% = 87.0 %

7. When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield,

and percent yield.

TiO2(s) + 2 C(s)➜Ti(s) + 2 CO(g)

smalleramount is

from limitingreactant}kg

g

g

mol mol

mol

kgC

kgTiO2

gC

gTiO2

molC

molTiO2

molTi

molTi

Theoretical Yield

smaller molTi

gTi

kgTi

Theoretical Yield

Actual Yield

= % Yield


and percent yield.

TiO2(s) + 2 C(s)➜Ti(s) + 2 CO(g)

Collect needed relationships:

1000 g = 1 kg Molar Mass Ti = 47.87 g/mol Molar Mass C = 12.01 g/mol

Molar Mass TiO2 = 79.87 g/mol


and percent yield.

TiO2(s) + 2 C(s)➜ Ti(s) + 2 CO(g)

1 mole TiO2 : 1 mol Ti 2 mole C : 1 mol Ti

88.2 kg TiO2 x x x

= 1.1043 x 103 mol Ti

limiting reactantTheoretical yield

7. When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield.

TiO2(s) + 2 C(s)➜Ti(s) + 2 CO(g)

1.10 x 103

28.6 kg C x x x

= 1.1907 x 103 mol Ti

1000 g C1 kg C

1.00 mol C12.01 g C

1.00 mol Ti2.00 mol C

1.19 x 103

1000 g TiO21 kg TiO2

1.00 mol TiO279.87 g TiO2

1.00 mol Ti1.00 mol TiO2

theoretical yield

percent yield


and percent yield.

TiO2(s) + 2 C(s)➜Ti(s) + 2 CO(g)

1.10 x 103 mol Ti x x = 52.9 kg Ti 47.87 g Ti1 mol Ti

1.00 kg Ti1000 g Ti

Enthalpy

A Measure of the Heat Evolved or Absorbed in a Reaction

Exothermic Reactions emit thermal energy when they occur.

Endothermic Reactions absorb thermal energy when they occur.

Enthalpy - The amount of thermal energy emitted or absorbed by a chemical reaction, under

conditions of constant pressure.

We can only measure the change in enthalpy, therefore the important quantity is ΔH.

Enthalpy

Molecular View of Exothermic Reactions

The temperature of the surroundings rises due to

release of thermal energy by the reaction.

The products of the reaction have less chemical potential energy than the reactants.

The difference in energy is released as heat to the

surroundings.

ΔH = Hf -Hi

ΔH is “-’’

REACTANTS

PRODUCTS

Hi

Hf

decrease in enthalpy

En

tha

lpy

Reaction coordinate

Molecular View of Endothermic Reactions

The temperature of the surroundings decreases due to absorption of thermal energy

by the reaction.

The products of the reaction have more chemical potential

energy than the reactants.

The difference in energy is absorbed and becomes part of the chemical potential energy

of the products. ΔH = Hf -Hi

ΔH is “+’’

REACTANTS

PRODUCTS

Hi

Hf

increase in enthalpy

En

tha

lpy

Reaction coordinate

Sign of ΔHrxn Combustion of CH4, the main component in natural gas:

This reaction is exothermic and therefore has a negative enthalpy of reaction.

The sign and magnitude of ΔHrxn tell us that 802.3 kJ of heat are emitted when 1 mol CH4 reacts with 2 mol O2.

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) + 802.3 kJ

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) ΔHrxn = -802.3 kJ

Sign of ΔHrxn

Reaction between nitrogen and oxygen gas to form nitrogen monoxide:

This reaction is endothermic and therefore has a positive enthalpy of reaction.

The sign and magnitude of ΔHrxn tell us that 182.6 kJ of heat are absorbed from the surroundings when 1 mol N2 reacts with 1 mol O2.

N2(g) + O2(g) → 2 NO(g) ΔHrxn = +182.6 kJ

182.6 kJ + N2(g) + O2(g) → 2 NO(g)

13.2 kg C3H8 x x x

= 6.1195 x 105 kJ

8. How much heat is evolved in the complete combustion of 13.2 kg of C3H8(g)?

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) ΔH = −2044 kJ

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) + 2044 kJ

kg C3H8 g C3H8 mol C3H8 kJ

1000 g

kg

1 molC3H8

44.09 g C3H8

2044kJ

1 molC3H8

1000 g C3H81.00 kg C3H8

1 mol C3H844.09 g C3H8

2044 kJ1 mol C3H8

6.12 x 105 kJ

ch1410 lecture #13 troch8 stoichiometry web...

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