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STOICHIOMETRY!. Homework  Page 285, #1-3  Review P. 288 – 295  P. 295, # 1-4 due Monday For tomorrow!

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Presentation on theme: "STOICHIOMETRY!. Homework  Page 285, #1-3  Review P. 288 – 295  P. 295, # 1-4 due Monday For tomorrow!"— Presentation transcript:

1 STOICHIOMETRY!

2 Homework  Page 285, #1-3  Review P. 288 – 295  P. 295, # 1-4 due Monday For tomorrow!

3 Stoichiometry!  Composition stoichiometry –  Reaction stoichiometry –

4 Mole Ratios  In a nutshell: How many moles of one (product or reactant) will you need/receive if you are given one amount of another?  (confused? Let’s do an example!)   What is the molar ratio between magnesium and hydrochloric acid?

5 Mole Ratios  What are all possible mole ratios for the following?   How might these be useful?

6 Stoichiometry!  Ratios of moles in a chemical equation can be used as conversion factors  In the above reaction, how many grams of magnesium will you need to react with 2.00 moles of hydrochloric acid?

7 Practice!  Determine all of the possible mole ratios for each of the following unbalanced reactions. ___Ba(ClO 3 ) 2  ___BaCl 2 + ___O 2 ___Ca + ___H 2 O  ___Ca(OH) 2 + ___H 2 ___MgCl 2  ___Mg + ___Cl 2 ___ Fe + ___Cu(NO 3 ) 2  ___Fe(NO 3 ) 2 + ___Cu ___KOH + ___H 2 SO 4  ___K 2 SO 4 + ___H 2 O ___Mg + ___O 2  ___MgO ___AgNO 3 + ___NaCl  ___AgCl + ___NaNO 3 ___C 2 H 6 + ___O 2  ___CO 2 + ___H 2 O

8 Stoichiometry  These should be set up as unit conversions!  These types are extremely important for determining the amount of product you should have at the end of an experiment  Let’s practice!

9 Practice  You have grams of iron (II) sulfide, which you use as a reactant in the following formula equation: FeS + HCl  FeCl 2 + H 2 S  How much hydrochloric acid is needed in this reaction? Give your answer in grams.  How much of each product is produced (in grams)?  When methane combusts, there are grams of oxygen gas available. How much methane is consumed in the reaction? How much of each product is produced (in grams)?

10 Homework  P.295, #1-5 – due in three school days

11 Limiting Reactants  Limiting reactant –  Excess reagent –  Limiting reactants occur whenever you do NOT have an ideal stoichiometric calculation.

12 Limiting Reactants  Step 1 – convert everything to moles  Step 2 – balance formula equation  Step 3 – compare molar values to the balanced chemical equations  Step 4 – determine the limiting reactant!  Step 5 – calculate the amount of the second reactant needed  Step 6 – subtract the amount of excess reactant needed from the excess amount available (this is your excess!)  Step 7 – determine the amount of products produced based on your limiting reactant (use stoichiometry!)

13 Limiting Reactant - example  Silicon dioxide (quartz) is usually quite unreactive, but readily reacts with hydrogen fluoride according to the following equation. SiO 2 (s) + HF(g)  SiF 4 (g) +2H 2 O (l)  If 6.0 moles of hydrogen fluoride are added to 4.5 moles of silicon dioxide, which is the limiting reactant?

14 More practice – class work  Some rocket engines use a mixture of hydrazine, N 2 H 4, and hydrogen peroxide, H 2 O 2, as the propellant. The formula equation is given by the following equation. N 2 H 4 (l) + H 2 O 2 (l)  N 2 (g) + 4H 2 O (g)  Which is the limiting reactant in this reaction, when moles of N 2 H 4 is mixed with moles of hydrogen peroxide?  How much of the excess reactant, in moles, remains unchanged?  How much of each product, in moles, is formed?

15 More Practice!  The black oxide of iron, Fe3O4, occurs in nature as the mineral magnetite. This substance can also be made in the laboratory by the reaction between red-hot iron and steam according to the following equation. Fe(s) + H 2 O (g)  Fe 3 O 4 (s) +H 2 (g)  When 36.0 g of steam are mixed with 67.0 grams of iron, which is the limiting reactant?  What mass in grams of black iron oxide is produced?  What mass in grams of excess reactant remains when the reaction is completed?

16 Percent Yield  Percent Yield  Example:  Chlorobenzene, C 6 H 5 Cl, is used in the production of many important chemicals, such as aspirin, dyes, and disenfectants. One industrial method of preparing chlorobenzene is to react benzene, C 6 H 6, with chlorine, as represented by the following equation. C6H6(l) + Cl2 (g)  C 6 H 5 Cl (l) + HCl (g)  When 36.8 grams of benzene reacts with excess chlorine gas, the actual yield of chlorobenzene is 38.8 grams. What is the percentage yield of chlorobenzene? Step 1: balance equation Step 2: Determine molecular weights Step 3: Determine theoretical yield Step 4: Determine percent yield = 73.2 %

17 Practice!  #3 on page 302


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