Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 11 and supplemental Ms. Lockhart Chemistry

Similar presentations


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

1 Chapter 11 and supplemental Ms. Lockhart Chemistry
Stoichiometry Chapter 11 and supplemental Ms. Lockhart Chemistry

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

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

4 Stoichiometry You can interpret a lot from a balanced chemical reaction. Iron is added to oxygen gas (O2) to yield iron (III) oxide (ferric oxide). 4 Fe O2  2Fe2O3 Atoms Moles Mass Total mass

5 Stoichiometry Nitrogen gas (N2) is added to hydrogen gas (H2) to make ammonia (NH3). N H2  NH3 Atoms Moles Mass Total mass

6 Practice Stoichiometry
3.00 g Magnesium (Mg) is heated with excess oxygen gas (O2) to make magnesium oxide. 2 Mg O2  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 O2  MgO

8 Steps to Stoichiometry
Write given information under the balanced chemical reaction 2 Mg O2  MgO 3.00 g

9 Steps to Stoichiometry
Write what you are looking for. 2 Mg O2  MgO 3.00 g ? O2

10 Steps to Stoichiometry
Convert given information into moles. 2 Mg O2  MgO 3.00 g ? O2 3.00 g Mg x 2 moles Mg = 3x2/24.31 = moles Mg 24.31 g Mg

11 Steps to Stoichiometry
Use mole ratio to get moles to the correct representative particles. 2 Mg O2  MgO 3.00 g ? O2 0.247 moles 0.247 moles Mg x 1 mole O2 = 0.247x1/2 = moles O2 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 O2  MgO 3.00 g ? O2 0.247 moles moles 0.124 moles O2 x 32 g O2 = 0.124x32/1 = 3.97 g O2 1 moles O2

13 Practice Stoichiometry
Correct significant figures (sig figs) and rewrite the answer. 3.00 g Mg = 3 S.F. S.F. Answer: 3.97 g O2

14 Let’s Take It Slow…


Download ppt "Chapter 11 and supplemental Ms. Lockhart Chemistry"

Similar presentations


Ads by Google