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04/17 Kinda like a recipe 4 Eggs + 3 Bacon  2 Omelets 8 Eggs + ? Bacon  ? Omlets  Al + 3O 2  2 Al 2 O 3 8 Al+? O 2  ? Al 2 O 3 +  + 

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Presentation on theme: "04/17 Kinda like a recipe 4 Eggs + 3 Bacon  2 Omelets 8 Eggs + ? Bacon  ? Omlets  Al + 3O 2  2 Al 2 O 3 8 Al+? O 2  ? Al 2 O 3 +  + "— Presentation transcript:

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2 04/17

3 Kinda like a recipe 4 Eggs + 3 Bacon  2 Omelets 8 Eggs + ? Bacon  ? Omlets  Al + 3O 2  2 Al 2 O 3 8 Al+? O 2  ? Al 2 O 3 +  + 

4 3 Stoichiometry (more working with ratios) Ratios are found within a chemical equation.  Al + 3O 2  2 Al 2 O 3 4 moles of aluminum react with 3 moles of oxygen to form 2 moles of aluminum oxide coefficients give MOLAR RATIOS

5 Mole to mole conversions  Al + 3O 2  2 Al 2 O 3 every time we react 3 moles of O 2 we make 2 moles of Al 2 O 3 2 moles Al 2 O 3 3 mole O 2 or 2 moles Al 2 O 3 3 mole O 2

6 Mole to Mole conversions How many moles of O 2 are produced when 3.34 moles of Al 2 O 3 decompose?  Al + 3O 2  2 Al 2 O 3 3.34 moles Al 2 O 3 2 moles Al 2 O 3 3 mole O 2 =5.01 moles O 2

7 6 When N 2 O 5 is heated, it decomposes: 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) a. How many moles of NO 2 can be produced from 4.3 moles of N 2 O 5 ? = moles NO 2 4.3 mol N2O5N2O5 8.6 b. How many moles of O 2 can be produced from 4.3 moles of N 2 O 5 ? = mole O 2 4.3 mol N2O5N2O5 2.2 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 4.3 mol? mol 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 4.3 mol ? mol Mole – Mole Conversions Units match

8 Formation of Ammonia…other ways to think about this

9 36.04 g reactants In terms of mass – Law of Conservation 2H 2 + O 2   2H 2 O 2 moles H 2 2.02 g H 2 1 moles H 2 = 4.04 g H 2 1 moles O 2 32.00 g O 2 1 moles O 2 = 32.00 g O 2 What should be the mass of the products?

10 36.04 g reactants 36.04 g products In terms of mass – Law of Conservation 2H 2 + O 2   2H 2 O 2 moles H 2 O 18.02 g H 2 O 1 mole H2O = 36.04 g H 2 O 2H 2 + O 2   2H 2 O Try using stoichiometry to get the products mass.

11 Practice! Show your work. Have your periodic table out. A. What is the mole ratio of N2O5 to O2 B. How many moles of oxygen will be produced from 3 moles of N2O5? 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g)

12 Practice! Show your work. Have your periodic table out. A. What is the mole ratio of N2O5 to O2 2:1 B. How many moles of oxygen will be produced from 3 moles of N2O5? 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) = 1. 5 Mol O 2 = 2 mol O2 3 mol N 2 O 5 2 mol N 2 O 5 1 mol O2

13 Periodic Table Moles A Moles B Mass g B Periodic Table Balanced Equation Mass g A Decide where to start based on the units you are given and stop based on what unit you are asked for

14 THE STEPS for Stoichiometry READ the problem and UNDERLINE what you know and need to find out WRITE known and what you want to know under equation SET UP house with known and what you want to know FIND pathway on green sheet to get from known to what you want to know Multiply/Divide and CANCEL units – What you want goes on top of the table!!!! RECORD answer (sig figs and units)

15 14 When N 2 O 5 is heated, it decomposes: 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) a. How many moles of N 2 O 5 were used if 210g of NO 2 were produced? = moles N 2 O 5 210 g NO 2 2.28= 2.3 b. How many grams of N 2 O 5 are needed to produce 75.0 grams of O 2 ? = grams N 2 O 5 75.0 g O2O2 506 gram ↔ mole and gram ↔ gram conversions 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 210g? moles 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 75.0 g ? grams Units match

16 15 Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? First write a balanced equation. Al(s) + HCl(aq)  AlCl 3 (aq) + H 2 (g)2 6 2 3 Gram to Gram Conversions

17 16 Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Al(s) + HCl(aq)  AlCl 3 (aq) + H 2 (g) 2 6 2 3 Now let’s get organized. Write the information below the substances. 3.45 g ? grams Gram to Gram Conversions

18 17 Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Al(s) + HCl(aq)  AlCl 3 (aq) + H 2 (g)2 6 2 3 3.45 g ? grams Let’s work the problem. = g AlCl 3 3.45 g Al We must always convert to moles.Now use the molar ratio.Now use the molar mass to convert to grams. 17.0 Units match gram to gram conversions

19 More Examples To make silicon for computer chips they use this reaction SiCl 4 + 2Mg  2MgCl 2 + Si How many moles of Mg are needed to make 9.3 g of Si? 3.74 mol of Mg would make how many moles of Si? How many grams of MgCl 2 are produced along with 9.3 g of silicon?

20 For Example The U. S. Space Shuttle boosters use this reaction 3 Al(s) + 3 NH 4 ClO 4  Al 2 O 3 + AlCl 3 + 3 NO + 6H 2 O How much Al must be used to react with 652 g of NH 4 ClO 4 ? How much water is produced? How much AlCl 3 ?


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