STOICHIOMETRY. I CAN solve a stoichiometry (mass – mass) problem using a chemical equation and...

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STOICHIOMETRY

Transcript of STOICHIOMETRY. I CAN solve a stoichiometry (mass – mass) problem using a chemical equation and...

Page 1: STOICHIOMETRY. I CAN solve a stoichiometry (mass – mass) problem using a chemical equation and mass data.

STOICHIOMETRY

Page 2: STOICHIOMETRY. I CAN solve a stoichiometry (mass – mass) problem using a chemical equation and mass data.

• I CAN solve a stoichiometry (mass – mass) problem using a chemical equation and mass data.

Page 3: STOICHIOMETRY. I CAN solve a stoichiometry (mass – mass) problem using a chemical equation and mass data.

What is Stoichiometry?

• Stoichiometry is the quantitative study of reactants and products in a chemical reaction.

Page 4: STOICHIOMETRY. I CAN solve a stoichiometry (mass – mass) problem using a chemical equation and mass data.

What You Should Expect To Do?

• Given : Amount of reactant(s)

• Question: How much product(s) can be formed?

• Example • X A + Y B ? C

• “Given X grams of A and Y grams of B, how many grams of C can be produced?”

Page 5: STOICHIOMETRY. I CAN solve a stoichiometry (mass – mass) problem using a chemical equation and mass data.

What Skills Do You Need?

Some skills you will need to use:

• Molar Ratios (come from coefficients of a BALANCED equation).• Molar Masses (Formula Weight calculations).

• Balancing Equations

• Conversions between grams and moles

Note: This type of problem is often called "mass-mass."

Page 6: STOICHIOMETRY. I CAN solve a stoichiometry (mass – mass) problem using a chemical equation and mass data.

Steps Involved in Solving Mass-Mass Stoichiometry Problems

• Balance the chemical equation correctly.

• Convert GRAMS to MOLES of the starting substance.

• Set up a MOLE RATIO between starting and ending substances.

• Convert MOLES to GRAMS of ending substance.

Page 7: STOICHIOMETRY. I CAN solve a stoichiometry (mass – mass) problem using a chemical equation and mass data.

Mole Ratios

A mole ratio COMPARES moles of one compound in a balanced chemical equation to moles of another compound (look at their coefficients).

Page 8: STOICHIOMETRY. I CAN solve a stoichiometry (mass – mass) problem using a chemical equation and mass data.

ExampleIn the reaction between magnesium and oxygen to form magnesium oxide, what

is the mole ratio for the substances?

2 Mg + O2 2 MgO

2 moles : 1 mole : 2 moles

Page 9: STOICHIOMETRY. I CAN solve a stoichiometry (mass – mass) problem using a chemical equation and mass data.

Steps to Solve a Stoichiometry Problem

• Make sure the equation is BALANCED.

• Label the QUANTITIES GIVEN.• DRAW a DIMENSIONAL ANALYSIS • FRAME:

Grams starting

X FW starting, g

1 molestarting

Moles starting

Moles ending

1 mole ending

FW ending, g

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• Fill in the frame with the MOLES and FWs as indicated on the frame.

• Complete the math across the top of the frame, then across the bottom.

• Finally divide the TOP by BOTTOM and ROUND to SIGNIFICANT FIGURES.

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Practice Problem

• If 75.2 g of KClO3 decomposes, how many grams of O2 will be produced?

2 KClO3 2 KCl + 3 O2

75.2 g ? g

✔balanced✔balanced

Page 12: STOICHIOMETRY. I CAN solve a stoichiometry (mass – mass) problem using a chemical equation and mass data.

75.2g KClO3 =

FW KClO3

XX 122.55 g KClO3

122.55 g KClO3

1 moleKClO3

1 moleKClO3

2 moleKClO3

2 moleKClO3

3 moleO2

3 moleO2

1 moleO2

1 moleO2

31.998 gO2

31.998 gO2

FW O2FW O2

MoleRatioMoleRatio

______ g O2______ g O2

Page 13: STOICHIOMETRY. I CAN solve a stoichiometry (mass – mass) problem using a chemical equation and mass data.

75.2g KClO3 =

XX 122.55 g KClO3

122.55 g KClO3

1 moleKClO3

1 moleKClO3

2 moleKClO3

2 moleKClO3

3 moleO2

3 moleO2

1 moleO2

1 moleO2

31.998 gO2

31.998 gO2

7218.7488 g O2

____________________

245.1

7218.7488 g O2

____________________

245.1

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=29.45225 g ≈ 29.5 g O2

=29.45225 g ≈ 29.5 g O2

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MORE PRACTICE

• #1) If 75.2 grams of Calcium Sulfate are reacted with an unlimited amount of Sodium Phosphate, what mass of Calcium Phosphate will be formed according to the reaction:

• CaSO4 + Na3PO4 Ca3(PO4)2 + Na2SO4

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• #2) How many grams of Potassium are needed to form 125.6 grams of Potassium Carbonate according to the reaction below?

• K + Li2CO3 K2CO3 + Li