STOICHIOMETRY: tells relative amts of reactants & products in a chemical reaction

• Given an amount of a substance involved in a chemical reaction, we can figure out the amount of the other substances are needed or produced
• Always compare the number of MOLES
• 4 Fe + 3 O₂ à 2 Fe₂O₃
• If I have 4 moles of Fe, I need 3 moles of O₂ in order to produce 2 moles of Fe₂O₃.
• Comparison of coefficients = MOLE RATIO
• In the above eqn, what is the mole ratio between Fe₂O₃ and O₂?
• Use coefficients… 2 : 3 or 2 to 3

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To solve stoichiometry problems… ALWAYS!!!!!!!!!!!

** WRITE THE BALANCED EQN & GIVEN INFORMATION! **

1.) Find moles of given element or compound.

* Use molar mass of given substance, if problem gives you grams.

2.) Use mole ratio (coefficients) from balanced equation.

3.) Find answer.

* Use molar mass of unknown substance, if question asks for grams.

So… here’s an example:
If 34.32 grams of acetylene (C₂H₂) are burned in air, how many moles of CO₂ can be formed?

2 C₂H₂ + 5 O₂ à 4 CO₂ + 2 H₂O

34.32 g ? moles

1.) 34.32 g C₂H₂ | 1 mole C₂H₂ = 1.32 moles C₂H₂ C: 2 x 12.0 = 24.0

| 26 grams H: 2 x 1.0 = 2.0 +

26.0

2.) 1.32 moles C₂H₂ = x moles CO₂ 2 x = 5.28
2 4 x = 2.64 moles CO₂

Problem asked for moles of CO₂, so that is your answer. Be sure to round for significant figures.
Answer should be reported as 2.640 moles CO₂

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Example 2:

How many grams of Na₂CO₃ are needed to react completely with 6.21 moles of Ca(OH)₂?

Na₂CO₃ + Ca(OH)₂ à 2 NaOH + CaCO₃

? g 6.21 moles

1.) Step 1 is already done for you! The number of moles is given in the problem! YAY!

2.) 6.21 moles Ca(OH)₂ = x moles Na₂CO₃ x = 6.21 moles Na₂CO₃

1 1

3.) 6.21 moles Na₂CO₃ | 106 grams = 658 g Na₂CO₃

| 1 mole

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Example 3:

If 27.8 grams of Na₂SO₄ are reacted with excess Al₂(CO₃)₃, how many grams of Al₂(SO₄)₃ will be formed?

3 Na₂SO₄ + Al₂(CO₃)₃ à Al₂(SO₄)₃ + 3 Na₂CO₃

27.8 g ? g

1.) 27.8 g Na₂SO₄ | 1 mole Na₂SO₄ = 0.196 moles Na₂SO₄ Na: 2 x 23.0 = 46.0

| 142.1 grams S: 1 x 32.1 = 32.1

O: 4 x 16.0 = 64.0 +

2.) 0.196 moles Na₂SO₄ = x mole Al₂(SO₄)₃ 142.1
3 1 3 x = 0.196
x = 0.0653 moles Al₂(SO₄)₃

3.) 0.0653 mole Al₂(SO₄)₃ | 342.3 g Al₂(SO₄)₃ = 22.4 g Al₂(SO₄)₃ Al: 2 x 27.0 = 54.0 | 1 mole Al₂(SO₄)₃ S: 3 x 32.1 = 96.3

O: 12 x 16.0 = 192.0 + 342.3

Practice Problem:

How many grams of calcium nitrate are formed when 57.9 grams of iron (III) nitrate react with excess calcium hydroxide according to the following equation?

___ Ca(OH)₂ + ___ Fe(NO₃)₃ à ___ Ca(NO₃)₂ + ___ Fe(OH)₃

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