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Stoichiometry Chapter 11 and supplemental Ms. Lockhart Chemistry.

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Presentation on theme: "Stoichiometry Chapter 11 and supplemental Ms. Lockhart Chemistry."— Presentation transcript:

1 Stoichiometry Chapter 11 and supplemental Ms. Lockhart Chemistry

2 Stoichiometry Importance By understanding the mole we can use the information to By understanding the mole we can use the information to Predict mass or volume of a product. Predict mass or volume of a product. Predict how much reactant is needed for a reaction. Predict how much reactant is needed for a reaction. Make substances of all kinds like Make substances of all kinds like Your sneakers Your sneakers Glue Glue Your clothes Your clothes Gum Gum Candy Candy Petrochemicals (oils, vaseline, gas) Petrochemicals (oils, vaseline, gas)

3 Other uses % by mass (% composition) % by mass (% composition) Empirical Formula Empirical Formula Formula of a Hydrate Formula of a Hydrate Stoichiometric calculations Stoichiometric calculations Limiting Reagent Limiting Reagent % yield % yield

4 Stoichiometry You can interpret a lot from a balanced chemical reaction. Iron is added to oxygen gas (O 2 ) to yield iron (III) oxide (ferric oxide). Iron is added to oxygen gas (O 2 ) to yield iron (III) oxide (ferric oxide). 4 Fe + 3 O 2 2Fe 2 O 3 4 Fe + 3 O 2 2Fe 2 O 3AtomsMolesMass Total mass

5 Stoichiometry Nitrogen gas (N 2 ) is added to hydrogen gas (H 2 ) to make ammonia (NH 3 ). Nitrogen gas (N 2 ) is added to hydrogen gas (H 2 ) to make ammonia (NH 3 ). N H 2 2 NH 3 N H 2 2 NH 3AtomsMolesMass Total mass

6 Practice Stoichiometry 3.00 g Magnesium (Mg) is heated with excess oxygen gas (O 2 ) to make magnesium oxide g Magnesium (Mg) is heated with excess oxygen gas (O 2 ) to make magnesium oxide. 2 Mg + O 2 2 MgO

7 Steps to Stoichiometry Copy the word equation (write the chemical reaction in words) Write the word equation into a formula equation Balance the chemical reaction. 2 Mg + O 2 2 MgO

8 Steps to Stoichiometry Write given information under the balanced chemical reaction 2 Mg + O 2 2 MgO 3.00 g

9 Steps to Stoichiometry Write what you are looking for. 2 Mg + O 2 2 MgO 3.00 g ? O 2

10 Steps to Stoichiometry Convert given information into moles. 2 Mg + O 2 2 MgO 3.00 g ? O g Mg x 2 moles Mg = 3x2/24.31 = moles Mg g Mg g Mg

11 Steps to Stoichiometry Use mole ratio to get moles to the correct representative particles. 2 Mg + O 2 2 MgO 3.00 g ? O moles moles Mg x 1 mole O 2 = 0.247x1/2 = moles O 2 2 moles Mg 2 moles Mg

12 Steps to Stoichiometry Convert mole to grams, particles or whatever the answer ask for Calculate and write the answer. 2 Mg + O 2 2 MgO 3.00 g ? O moles moles moles O 2 x 32 g O 2 = 0.124x32/1 = 3.97 g O 2 1 moles O 2 1 moles O 2

13 Practice Stoichiometry Correct significant figures (sig figs) and rewrite the answer g Mg = 3 S.F g Mg = 3 S.F. S.F. Answer: 3.97 g O 2 S.F. Answer: 3.97 g O 2

14 Lets Take It Slow…


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