stoichiometry. i can solve a stoichiometry (mass – mass) problem using a chemical equation and...

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.

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?”

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."

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.

Mole Ratios

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

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

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

• 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.

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

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

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

=29.45225 g ≈ 29.5 g O2

=29.45225 g ≈ 29.5 g O2

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

• #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