Chapter 12 Stoichiometry part 1. Stoichiometry The study of quantitative relationships between...

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Chapter 12 Chapter 12 Stoichiometry Stoichiometry part 1 part 1

Transcript of Chapter 12 Stoichiometry part 1. Stoichiometry The study of quantitative relationships between...

Chapter 12 Chapter 12 StoichiometryStoichiometry

part 1part 1

StoichiometryStoichiometry

The study of quantitative The study of quantitative relationships between amounts of relationships between amounts of

reactants used and products formed reactants used and products formed by a chemical reactionby a chemical reaction

StoichiometryStoichiometry

• Calculations of quantities in chemical reactions

• This means using balanced equations to calculate quantities of chemicals used in a chemical reaction

Law of Conservation of MassLaw of Conservation of Mass

• Matter is neither created nor destroyed in a chemical reaction

• Chemical bonds in reactants break and new chemical bonds form to produce products

• Mass of reactants equals the mass of the products

A chemical equation is like a A chemical equation is like a recipe; it shows you how much recipe; it shows you how much of each “ingredient” is used and of each “ingredient” is used and how much product you will have how much product you will have

in the end.in the end.

The heart of stoichiometry is using a balanced chemical equation as a conversion

factor for dimensional analysis.

NN22 + 3H + 3H22 2NH 2NH33

• How many molecules of N2 are there?

• How many molecules of H2 are there?

• How many molecules of NH3 are there?

So one molecule of N2 will react with 3 molecules of H2 to form 2 molecules of NH3.

You could also say that one mole of N2 will react with 3 moles of H2 to form 2 moles of NH3.

• This equation tells the proportions of N2, H2, and ammonia

• The coefficients are in a 1:3:2 ratio

• The ratio could mean molecules or moles of each

Review

6.02 x 1023 particles

1 mole

molar mass

1N2(g) + 3H2(g) 2NH3(g)

1 mole 3 moles 2 moles

28.00g 3(2.02)=6.06g 2(14.00+3.03)=34.06g

Steps to solve every problem:Steps to solve every problem:

1) Write a balanced equation.2) Put what you have over 1.3) Convert into moles.4) Compare using the equation.5) Convert into the unit the

question asks for.

Mole to Mass conversion:Mole to Mass conversion:

The mole ratios of the balanced equation is known, and you’re given a quantity in moles, now find the

mass of another product or reactant.

Mole to Mole conversions:

Conversion factor (as with any conversion factor, units must cancel)

Moles of known x moles of unknown = moles unknown Moles of known

Burning of propane gas:Burning of propane gas:C3H8(g) + 5O2(g) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) 3CO2(g) + 4H2O(g)

How much oxygen is needed to produce 18 mol of CO2?

Zn + HCl Zn + HCl ZnCl ZnCl22 + H + H22

• For the above reaction, how many moles of zinc chloride can be produced with 0.27 moles of zinc?

• For the above reaction, how many moles of zinc will react with 1.39 moles of hydrochloric acid?

Fe + OFe + O22 Fe Fe22OO33

• How many grams of iron (III) oxide can be produced with 17.2g Fe?

• How many molecules of O2 are needed to produce 3.92g Fe2O3?

Mass to Mass conversion: Mass to Mass conversion: Again, the mole ratios of the balanced

equation is known and you’re given a quantity in grams, now find the mass of

another product or reactant.

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

67.02g propane would produce ? grams of water vapor?