proportional relationships stoichiometrystoichiometry –mass relationships between substances in a...

Proportional Relationships

• StoichiometryStoichiometry– mass relationships between substances in a chemical

reaction– based on the mole ratio

• Mole RatioMole Ratio– indicated by coefficients in a balanced equation

2 Mg + O2 Mg + O22 2 MgO 2 MgO

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Stoichiometry Island Diagram

Mass

Particles

Mole Mole

Mass

Particles

Known Unknown

Substance A Substance B

Stoichiometry Island Diagram

1 mole = molar mass (g) Use coefficientsfrom balanced

chemical equation

1 mole =

6.022 x

1023 partic

les

(atoms o

r molecu

les)1 mole = 6.022 x 10 23 particles

(atoms or molecules)

1 mole =

molar m

ass (g

)

?

Visualizing a Chemical Reaction

Na + Cl2 NaCl

___ mole Cl2 ___ mole NaCl___ mole Na

2

10 5 10

2

10 5 10

Formation of Ammonia

Stoichiometry Steps1. Write a balanced equation.2. Identify known & unknown.3. Line up conversion factors.

– Mole ratio - moles moles– Molar mass - moles grams– Avogadro’s number - particles moles

Core step in all stoichiometry problems!!

– Mole ratio - moles moles

4. Check answer.Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Stoichiometry Problems• How many moles of KClO3 must decompose

in order to produce 9 moles of oxygen gas?

9 mol O2 2 mol KClO3

3 mol O2

= 6 mol KClO3

2KClO3 2KCl + 3O2 ? mol 9 mol


Stoichiometry Problems• How many grams of silver will be formed

from 12.0 g copper?

12.0g Cu

1 molCu

63.55g Cu

= 40.7 g Ag

Cu + 2AgNO3 2Ag + Cu(NO3)2

2 molAg

1 molCu

107.87g Ag

1 molAg

12.0 g ? g


Rocket Fuel The compound diborane (B2H6) was at one time considered for use as a rocket fuel. How many grams of liquid oxygen would a rocket have to carry to burn 10 kg of diborane completely? (The products are B2O3 and H2O).

B2H6 + O2Chemical equation

Balanced chemical equation

X = 34,286 g O2

10 kg x g

x g O2 = 10 kg B2H6

1000 g B2H6

1 kg B2H6 28 g B2H6

1 mol B2H6 3 mol O2

1 mol B2H6

32 g O2

1 mol O2

B2O3 + H2O

3 3B2H6 + O2 B2O3 + H2O

Limiting Reactants

• Limiting ReactantLimiting Reactant– used up in a reaction– determines the amount of product

• Excess ReactantExcess Reactant– added to ensure that the other reactant is

completely used up– cheaper & easier to recycle


Percent Yield

calculated on paper

measured in lab

% yield =actual yield

theoretical yieldx 100

When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl

are formed. Calculate the theoretical and % yields of KCl.

K2CO3 + 2KCl + H2O + CO2 45.8 g

46.3 g

actual yield

excess

2HCl

theoretical yield

Theoretical yield

x g KCl = 45.8 g K2CO3 = 49.4 g KCl1 mol K2CO3

138 g K2CO3

2 mol KCl1 mol K2CO3

46.3 g KCl% Yield =

Actual Yield

Theoretical Yield% Yield

% Yield = 93.7% efficient

49.4 g KCl

= x 100

74.5 g KCl1 mol KCl

49.4 g

? g

proportional relationships stoichiometrystoichiometry –mass relationships between substances in a...

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