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Section 2.9 page 144 Stoichiometry is the science of using balanced chemical equations to determine exact amounts of chemicals needed or produced in a.

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Presentation on theme: "Section 2.9 page 144 Stoichiometry is the science of using balanced chemical equations to determine exact amounts of chemicals needed or produced in a."— Presentation transcript:

1 Section 2.9 page 144 Stoichiometry is the science of using balanced chemical equations to determine exact amounts of chemicals needed or produced in a chemical reaction.

2 Procedure for solving Stoichiometric problems Example: Propane, C 3 H 8 (g), is a gas that is commonly used in barbecues. Calculate the mass of oxygen that is needed to burn 15 g of propane. 1. Write the balance equation: 2. Write out the ratio. (represents the # of moles needed)

3 Propane, C3H8(g), is a gas that is commonly used in barbecues. Calculate the mass of oxygen that is needed to burn 15 g of propane. 3. Indicate the known substances mass 4. Convert known substances mass into moles (you need the known mass and the molar mass)

4 Propane, C3H8(g), is a gas that is commonly used in barbecues. Calculate the mass of oxygen that is needed to burn 15 g of propane 5. Set up a proportion of from the ratio of # mole required and # of moles calculated, from given substance and unknown. 6. Solve for moles of unknown substance by solving the proportion above

5 Propane, C3H8(g), is a gas that is commonly used in barbecues. Calculate the mass of oxygen that is needed to burn 15 g of propane 7. Using the calculated unknown moles solve for grams using the # moles and Molar mass of unknown.

6 How many molecules of oxygen are produced from the decomposition of 12 g of water into its elements?

7 How many grams of oxygen are required to react with 9.70 g of magnesium to form magnesium oxide?

8 Potassium chlorate breaks down to form potassium chloride and oxygen gas. What mass of potassium chloride must be decomposed to form 0.96 g of oxygen?

9 Assignment Page 148 questions 1 and 2 1. 2 Al 2 O 3  4 Al + 3 O 2 2. 2 K + 2 HCl  2 KCl + H 2


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