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1 Chemical Equation Describes a chemical reaction A + B  C A and B = reactants C = product CO (g) + 2H 2(g)  CH 3 OH (l)

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Presentation on theme: "1 Chemical Equation Describes a chemical reaction A + B  C A and B = reactants C = product CO (g) + 2H 2(g)  CH 3 OH (l)"— Presentation transcript:

1 1 Chemical Equation Describes a chemical reaction A + B  C A and B = reactants C = product CO (g) + 2H 2(g)  CH 3 OH (l)

2 2 Information Given by Chemical Equations CO (g) + 2H 2(g)  CH 3 OH (l) 1 molecule of CO + 2 molecules of H 2  one molecule of CH 3 OH 10 molecules of CO + 20 molecules of H 2  10 molecules of CH 3 OH 1 mole of CO + 2 moles of H 2  1 mole of CH 3 OH Do you notice a relationship pattern?

3 3 Stoichiometry Process of using a chemical equation to calculate the relative masses of reactants and products involved in a reaction.

4 4 Balancing equations In a chemical reaction, atoms are neither created nor destroyed. All atoms present in the reactants must be accounted for among the products. Therefore, one needs to balance the chemical equation for the reaction. What goes in must come out, albeit in a different form.

5 5 Rules for balancing equations 1. Balance out metals and polyatomic ions first Consider polyatomic ions as a single species 2. Balance out elements that are found in more than one species last Leave H for last: Na + H 2 O  NaOH + H 2 3. Don’t get discouraged! Balancing one atom can throw others off Balance the equation afresh 4. Do NOT change the nature of the compound To balance H 2 O you cannot make it H 2 O 2 Why not? 5. Thus, put numbers in front of compounds 6. Lowest whole numbers Incorrect: 4A + 4B  4AB Correct: A + B  AB

6 6 Example Balance: Na (s) + H 2 O (l)  NaOH (aq) + H 2(g) Na is balanced, O is balanced, but H is not  balance H  put 2 in front of NaOH This gives us 4 H on the products side Add 2 in front of water on reactants side H’s are now balanced − So are O’s (2 on each side) Na is now unbalanced Add a 2 to Na on reactant side Recapitulate: check equation 2Na (s) + 2H 2 O (l)  2NaOH (aq) + H 2(g) BALANCED!

7 7 Balance the following Al (s) + F 2(g)  AlF 3(s) KClO 3(s)  KCl (s) + O 2(g) KI (aq) + Pb(NO 3 ) 2(aq)  KNO 3(aq) + PbI 2(s) C 2 H 6(g) + O 2(g)  CO 2(g) + H 2 O (l) Let’s talk about a neat trick for the above

8 8 Mole-Mole Relationships: Dimensional Analysis Consider the decomposition of water: 2H 2 O (l)  2H 2(g) + O 2(g) If we decompose 4 mol of H 2 O, how many moles of O 2 do we get? If we decompose 5.8 mol of H 2 O, how many moles of H 2 do we get?

9 9 More practice Calculate the moles of C 3 H 8 used when 4.30 moles of CO 2 are obtained. C 3 H 8(g) + 5O 2(g)  3CO 2(g) + 4H 2 O (g)

10 10 Mass Calculations Steps for Calculating the Masses of Reactants and Products in Chemical Reactions 1. Balance the equation for the reaction. 2. Convert the masses of reactants or products to moles. 3. Use the balanced equation to set up the appropriate mole ratio(s). 4. Use the mole ratio(s) to calculate the number of moles of the desired reactant or product. 5. Convert from moles to mass.

11 11 In other words… grams A  moles A moles A  moles B moles B  grams B Do it all as a dimensional analysis problem! See next slide

12 12 Example What mass of  2 should we weigh out to react with 35.0 g Al? 2Al (s) + 3  2(s)  2Al  3(s) grams A  moles A : moles A  moles B : moles B  grams B : In one-step: Avoids rounding errors too!

13 13 Practice What mass of CO 2 is produced when 96.1 g of C 3 H 8 react with sufficient O 2. C 3 H 8(g) + 5O 2(g)  3CO 2(g) + 4H 2 O (g)

14 14 Calculations Involving a Limiting Reactant When there is insufficient amount of one of the reactants. Calculate the mass required for the product in question using both reactants. Which ever gives the least amount of product is the limiting reactant. The mass of product obtained from using the limiting reactant is the correct amount.

15 15 Example What mass of AlI 3 will we obtain if we react 35.0 g Al with g I 2 ? 2Al (s) + 3I 2(s)  2AlI 3(s) 35.0 g Al x (mol/ g) x (2 mol AlI 3 /2 mol Al) x ( g/mol) = 529 g AlI 3 Versus: g I 2 x (mol/ g I 2 ) x (2 mol AlI 3 /3 mol I 2 ) x ( g/mol) = 428 g AlI 3  I 2 is the limiting reactant. The mass of AlI 3 obtained is 428 g.

16 16 Practice Example Calculate the mass of lithium nitride formed from 56.0 g of N 2 and 56.0 g of Li in the balanced equation: 6Li (s) + N 2(g)  2Li 3 N (s)

17 17 More practice Example Calculate the mass of aluminum sulfate formed from 50.0 g of Al and 50.0 g of H 2 SO 4 in the balanced equation: 2Al (s) + 3H 2 SO 4(aq)  Al 2 (SO 4 ) 3(s) + 3H 2(g)

18 18 Percent Yield Theoretical Yield: Amount of product predicted from the amounts of reactants used. Actual Yield: Amount of product actually obtained. Actual yields are determined by doing the reaction __Actual Yield__ X 100% = Percent Yield Theoretical Yield

19 19 Percent Yield Consider the reaction: TiCl 4(g) + O 2(g)  TiO 2(s) + 2Cl 2(g) (a) Suppose 6.71 x 10 3 g of TiCl 4 is reacted with 2.45 x 10 3 g of O 2. Calculate the maximum mass of TiO 2 that can form. (b) If 2.12 x 10 3 g was produced in the laboratory, what is the percent yield?


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