Unit 8Stoichiometric Relationships In

Chemical ReactionsMole RatiosStoichiometry applied to gasesFractional Components of Chemical SamplesChemical Reactions Give Off or Take In EnergyLimiting And Excess Reagent Problems

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Mole Ratios

The coefficients of a chemical equation represent a ratio of moles. They are used to balance the equation thereby reflecting the Law of Conservation of Matter.

This law states that the a) total number of atoms during a chemical reaction does not change because matter cannot be created nor destroyed. b) atoms are rearranged.

Mole Ratios

For the equation2 H2 + 1 O2 ———> 2 H2O

2 moles of H2 react with 1 mole of O2 to get 2 moles of H2O.

4 +hydrogen atoms

2 yields oxygen atoms

2 water molecules

[6 atoms]

Furthermore, the Law of Conservation of Matter shows that

Q The salt potassium chlorate undergoes decomposition when heat is applied to form the salt potassium chloride and the gas oxygen.

a) Write the balanced equation.

b) What is the ratio of moles: potassium chlorate to potassium chloride to oxygen?

2 2 3

KClO3 KCl + O22 2 3

• The coefficients in the balanced chemical equation shows the molecules and mole ratio of the reactants and products

• Since moles can be converted to masses, we can determine the mass ratio of the reactants and products as well

Information Given by theChemical Equation

Stoichiometry Rules!

• Write a chemical equation.

• Balance the equation!

• Determine start and destination for dimensional analysis.

• Convert start details into moles.

• Multiply by the appropriate mole ratio.

• Convert to correct units for destination.

• Write the balanced equation 2 CO + O2 —> 2 CO2

• Use the coefficients to find the mole relationship 2 moles CO = 1 mol O2 = 2 moles CO2

Determine the Number of Moles of Carbon Monoxide required to react with 3.2 moles Oxygen,

and determine the moles of Carbon Dioxide produced

Example #1

Use dimensional analysis

CO moles 6.4O mole 1

CO moles 2xO moles 3.2

2 2

22

22 CO moles 6.4

O mole 1

CO moles 2 xO moles 3.2

Example #1Determine the Number of Moles of Carbon Monoxide

required to react with 3.2 moles Oxygen, and determine the moles of Carbon Dioxide produced

• Use the molar mass of the given quantity to convert it to moles

• Use the mole relationship to convert the moles of the given quantity to the moles of the desired quantity

Example #2Determine the Number of grams of Carbon Monoxide

required to react with 48.0 g Oxygen, and determine the mass of Carbon Dioxide produced

2

22

O mol 1CO mol 2

x g 32.00

O mol 1x O g 48.0

2

222

O mol 1CO mol 2

x g 32.00

O mol 1x O g 48.0

Use the molar mass of the desired quantity to convert the moles to mass

CO g 84.0CO mol 1g 28.01

x O mol 1CO mol 2

x g 32.00

O mol 1x O g 48.0

2

22

222

222 CO g 132

CO mol 1g 44.01

x O mol 1CO mol 2

x g 32.00

O mol 1x O g 48.0

Example #2Determine the Number of grams of Carbon Monoxide

required to react with 48.0 g Oxygen, and determine the mass of Carbon Dioxide produced

If 0.783 mole of potassium chlorate decomposed, how many moles of oxygen form?

If 2.04 grams of potassium chloride were produced, how many grams of potassium chlorate must have decomposed?

= mol O2 1.17 KClO mol 2

O mol 3

3

2

1

KClO mol 0.783 3

= g KClO33.35

KCl n 2

KClO n 2 3

1

KCl g 2.04

KCl g 74.6

KCl n 1

3

3

KClO n 1

KClO g 122.6

KClO3 KCl + O22 2 3

Limiting And Excess Reagent Problems

8 meals

cwx.prenhall.com

Determine the number of packaged meals based on the given supply:

3 Excess sandwiches

2 Excess oranges

One packaged meal for the USM service day:

In the chemical reaction

Mg(s) + 2 HCl(aq) ———> MgCl2(aq) + H2(g)

performed in many high school laboratories, more HCl is usually used than is needed. This is done to assure that all the magnesium reacts. The HCl, therefore, is in excess and some of it is left over when the reaction stops. No magnesium will be present once the reaction is over.

Product

+

8 C atomsor 8 mol

6 O moleculesor 6 mol

2 2 C atomsor 2 mol

6 CO moleculesor 6 mol

2

—>

For another chemical reaction C + O2 CO2

Initial supply of reactant

1. In many chemistry problems you are told which reactant is in excess and solve based on the specified quantities of the other reactant.

2. In some instances, however, you will be given quantities of both reactants. One is probably in excess, but you won’t be told which one.

3. The following problem shows one way to determine the reactant that is in excess

Q 5.00 g of calcium reacts with 5.00 g of sulfur in a composition reaction. Do a calculation to find thea) Excess reactant.

Ca(s) + S(s) CaS(s)

CaS n 0.125Ca n 1

CaS n 1

Ca g 40.1

Ca n 1

1

Ca g 5.00

CaS n 0.156S n 1

CaS n 1

S g 32.1

S n 1

1

S g 5.00

This means that S is in excess.

In other words, after the reaction there will be CaS,unreacted S, and no Ca.

Higher.

Q 5.00 g of calcium reacts with 5.00 g of sulfur in a composition reaction. Do a calculation to find theb) Grams of product formed.

CaS g 9.025CaS n 1

CaS g 72.2

Ca n 1

CaS n 1

Ca g 40.1

Ca n 1

1

Ca g 5.00

Start with the limiting reagent since all of it will be used up.

CaS g 11.24CaS n 1

CaS g 72.2

S n 1

CaS n 1

S g 32.1

S n 1

1

S g 5.00

Look what happens if you start with the excess reactant:

Ca(s) + S(s) CaS(s)

c) Calculate the grams of excess reactant unreacted once the reaction is completed.

10.0 g of original reaction mixture – 9.025 g of CaS formed

= 1.0 g S left over

More Practice4.00 g of calcium is mixed with 4.00 g of phosphorus

and ignited to get solid calcium phosphide, Ca3P2.

a) Determine which reactant is in excess. (ans. P)

b) How many grams of product were formed? (ans. 6.05 g)

c) How many grams of the excess reactant were left over once the reaction was completed? (8 grams reactants – 6.05 grams calcium phosphide = 1.95 grams phosphorus left)