01 StoichiometryChapter 12. What conversion factors would you need if you were going to move from grams to liters? Solve the following problems. –How.

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Presentation transcript:

01 StoichiometryChapter 12

What conversion factors would you need if you were going to move from grams to liters? Solve the following problems. –How many grams are there in 6 liters of CO2 –How many molecules are there in 4.2 grams of H20?

How do we use balanced equations? 02 Use balanced equations to figure out how much of a reactant we need or how much of a product is formed A balanced equation tells you how many moles of each reactant and product you have Use coefficient (numbers in front) The total number of moles of reactants DOES NOT equal the total number of moles of products

03 Mole Ratios A mole ratio shows the relationship between two “parts” of a balanced chemical equation. We can compare reactant to reactant How much of A do we need to react with 6 moles of B to create C? We can compare reactant to product How much of C will be produced from 5 moles of A? We can compare product to reactant How much of A do we need if we have 2 moles of C? Use this balanced equation to develop six mole ratios: N 2 + 3H 2  2NH 3

DEFAULT STYLES 04 Calculate the mole ratios Al + O 2  Al 2 O 3 #1.#2. #3.#4. #5.#6.

05 Mole to Mole Conversions Using the mole ratios we can do mole to mole conversions. We can use the following formula: 4Al + 3 O 2  2Al 2 O 3 How many moles of aluminum are needed to react with 6 moles of O 2 ? (starting with oxygen going to aluminum) 6 moles O 2 x 4Al = 3 O 2 How many moles of oxygen are needed to form 6 moles of Al 2 O 3 ?

Practice Problems 4Al + 3 O 2  2Al 2 O 3 How many moles of Oxygen are required to completely react with 25 moles of Aluminum? How many moles of Aluminum Oxide are produced from 16 moles of Oxygen with Aluminum?

Volume to Volume Calculations Using the mole ratios we can do volume to volume conversions. We can use the following formula: 4Al + 3 O 2  2Al 2 O 3 How many liters of aluminum are needed to react with 12.6 L of O 2 ? (starting with oxygen going to aluminum) L O 2 x 1 mole x 4Al x 22.4 L Al = 22.4 L 3 O 2 1 mole Al How many liters of oxygen are needed to form 32.6 L of Al 2 O 3 ?

Practice Problems N 2 (g) + 3 H 2 (g)  2 NH 3 (g) How many liters of Hydrogen are required to completely react with 225 liters of Nitrogen? How many liters of Ammonia (NH 3 ) are produced from 16 liters of Hydrogen?

Mass to Mass Conversions Using the mole ratios we can do mass to mass conversions. We can use the following formula: 4Al + 3 O 2  2Al 2 O 3 How many grams of aluminum are needed to react with 6.5 grams of O 2 ? (starting with oxygen going to aluminum) 6.5 grams O 2 x 1 mole O 2 x 4Al x grams Al = 32 grams 3 O 2 1 mole Al How many grams of oxygen are needed to form 22.9 g of Al 2 O 3 ?

Mass Practice Problems N 2 (g) + 3 H 2 (g) --> 2 NH 3 (g) How many grams of Hydrogen are required to completely react with 21.5 grams of Nitrogen? How many moles of Nitrogen are needed to react with grams of Hydrogen?