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STOICHIOMETRY IS THE PART OF CHEMISTRY THAT DEALS WITH THE AMOUNTS OF SUBSTANCES INVOLVED IN CHEMICAL REACTIONS. HENCE, IT IS THE STUDY OF THE QUANTITATIVE.

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Presentation on theme: "STOICHIOMETRY IS THE PART OF CHEMISTRY THAT DEALS WITH THE AMOUNTS OF SUBSTANCES INVOLVED IN CHEMICAL REACTIONS. HENCE, IT IS THE STUDY OF THE QUANTITATIVE."— Presentation transcript:

1 STOICHIOMETRY IS THE PART OF CHEMISTRY THAT DEALS WITH THE AMOUNTS OF SUBSTANCES INVOLVED IN CHEMICAL REACTIONS. HENCE, IT IS THE STUDY OF THE QUANTITATIVE RELATIONSHIPS IN CHEMICAL REACTIONS. A BALANCED CHEMICAL EQUATION INDICATES THE RELATIVE NUMBER OF MOLES OR “PARTICLES” INVOLVED IN A CHEMICAL REACTION. Stoichiometry

2 H 2 O (g) NO (g) O 2(g) NH 3(g) molecules5 molecules4 molecules6 molecules 4 moles5 moles4 moles6 moles g g g g Recall the definition of a mole: We can expand this definition to include gases. Early chemists primarily worked with gases, and devised many laws to explain the behavior(s) of these gases. g-formula mass = 1 mole = x particles

3 In 1811 Avogadro postulated his own gas law. He said that, “Equal volumes of gases contain equal numbers of molecules.” Thus, the volume is directly proportional to the number of particles of the number of moles. g-formula mass = 1 mole = x particles = 22.4 dm 3 at STP* *Standard Temperature and Pressure 0  C and 1.0 atm

4 H 2 O (g) NO (g) O 2(g) NH 3(g) liters5 liters4 liters6 liters By writing balanced chemical equations and incorporating mole conversions, you can calculate the amount of reactants needed or the amount of products produced. This is due to the fact that the equation indicates the number of moles of reactants and products.

5 TO SOLVE: 1.Write the balanced equation. 2.Find the number of moles of the given substance. 3.Inspect and use the balanced equation to determine the ratio of moles of given substance to moles of “wanted” substance. 4.Convert the amount of “wanted” substance into the desired units.

6 THE TYPICAL FORMAT: (given) grams1 mole (given)moles of wanted g-formula mass (wanted) g-formula mass (given)moles of given1 mole (wanted) Step#2 Step#3 from equation Step#4

7 Example: Sulfuric acid is neutralized by the addition of sodium hydroxide. How many grams of water are produced when 15. grams of sodium hydroxide react? TO SOLVE: 1.Write the balanced equation. 2.Find the number of moles of the given substance. 3.Inspect and use the balanced equation to determine the ratio of moles of given substance to moles of “wanted” substance. 4.Convert the amount of “wanted” substance into the desired units. H 2 SO 4(aq) + NaOH (aq) Na 2 SO 4(aq) + HOH (l) + H 2 O (l) g??? g 15 g NaOH1 mole NaOH g NaOH 2 mole H 2 O 2 mole NaOH g H 2 O 1 mole H 2 O = 6.8 g H 2 O

8 Example: How many moles of NaOH are required to neutralize 4.2 moles of H 2 SO 4 ? TO SOLVE: 1.Write the balanced equation. 2.Find the number of moles of the given substance. 3.Inspect and use the balanced equation to determine the ratio of moles of given substance to moles of “wanted” substance. 4.Convert the amount of “wanted” substance into the desired units. H 2 SO 4(aq) + NaOH (aq) Na 2 SO 4(aq) + H 2 O (l) 22 ??? moles 4.2 moles 4.2 moles H 2 SO 4 2 moles NaOH 1 mole H 2 SO 4 = 8.4 moles NaOH

9 Example: How many grams of Na 2 SO 4 are produced from 4.2 moles of H 2 SO 4 ? TO SOLVE: 1.Write the balanced equation. 2.Find the number of moles of the given substance. 3.Inspect and use the balanced equation to determine the ratio of moles of given substance to moles of “wanted” substance. 4.Convert the amount of “wanted” substance into the desired units. H 2 SO 4(aq) + NaOH (aq) Na 2 SO 4(aq) + H 2 O (l) moles ??? g 4.2 moles H 2 SO 4 1 mole Na 2 SO 4 1 mole H 2 SO gNa 2 SO 4 1 mole Na 2 SO 4 = g Na 2 SO 4 = 6.0 x 10 2 g Na 2 SO 4

10 So far you have worked stoichiometry problems in which the given quantity of a reactant is consumed completely and we use that quantity to figure out how much product is formed. However, in most instances you mix different amounts of various reactants and not all completely react. Lets look at the following hypothetical reaction: A + B C Start: 5 mol 3 mol 0 mol End: 2 mol 0 mol 3 mol As you can clearly see, all three moles of B were consumed but since you had an excess of A, 2 moles were left over.

11 The reagent that is consumed 100% is referred to as the limiting reagent or reactant (LR). The LR allows you to calculate (just like you learned previously) the amount of product formed and how much of the other reactant was consumed. This is a lot simpler to understand with an example.

12 Limiting reactant analogies…. For example, bicycles require one frame and two wheels. If you have 20 wheels but only 5 frames, it is clear that the number of frames will determine how many bicycles can be made.

13 TO SOLVE LIMITING REACTANT (REAGENT) PROBLEMS: 1.Write the balanced equation for the reaction. 2.Calculate the number of moles of each reactant. 3.Convert one of the reactants into the other. This tells you how much of each you need to complete the reaction. 4.Compare the moles calculated in step #2 to the amounts in #3. The one you have less of is the limiting reagent. 5.Use the limiting reagent as the given, and calculate the desired quantity.

14 EXAMPLE: The Haber Process is used to convert nitrogen and hydrogen gases to ammonia. If 20. g N 2 and 10. g H 2 are mixed, what is the limiting reagent? N 2(g) + H 2(g) NH 3(g) g 20. g 1.Write the balanced equation for the reaction. 2.Calculate the number of moles of each reactant. 20. g N 2 1 mole N g N g H 2 1 mole H g H 2 = 0.71 moles N 2(g) = 5.0 moles H 2(g)

15 3.Convert one of the reactants into the other. This tells you how much of each you need to complete the reaction. 4.Compare the given amounts to the answer in #3. The one you have less of is the limiting reagent. You have 5 moles of H 2 …..more than enough… N 2 is the limiting reagent. There is an excess of H 2, After the reaction has completed there will be no N 2 left, yet there will still be extra H 2. =.71 N 2 given and 2.1 moles need to react with it moles N 2 3 mole H 2 1 mole N 2 = 2.1 moles H 2 needed

16 EXAMPLE: How much NH 3 is produced? Use the L R (N2 in this problem) to determine amounts of products….. From the last example, we know N 2(g) + H 2(g) NH 3(g) moles N 2 2 mole NH 3 1 mole N g NH 3 1 mole NH 3 = 24 g NH 3

17 EXAMPLE: How much excess is there? Use the L R (N2 in this problem) to determine how much of the other reactant (H2 in this problem) is left over. From the last example, we know N 2(g) + H 2(g) NH 3(g) moles N 2 3 mole H 2 1 mole N gH 2 1 mole H 2 = g H 2 reacted 10. g H 2 to start g H 2 reacted g in excess = 6 g H 2 left over

18 EXAMPLE: How many grams of aluminum sulfide can form from the reaction of 9.00 g of aluminum with 8.00 g of sulfur? Al (s) + S (g) Al 2 S 3(s) g 9.00 g 9.00 g Al 1 mole Al g Al 8.00 g S1 mole S g S = moles Al (S) = moles S (s) 2 ??? g There is more than enough aluminum moles S2 moles Al 3 moles S = moles Al needed

19 Use the L R (S in this problem) to determine amounts of products… moles S 1 mole Al 2 S 3 3 mole S g Al 2 S 3 1 mole Al 2 S 3 = 12.5 g Al 2 S 3 produced Al (s) + S (g) Al 2 S 3(s) g 9.00 g 2 ??? g

20 EXAMPLE: How many grams of N 2 F 4 can theoretically be obtained from 4.00 g of NH 3 and 14.0 g of F 2 according to: NH 3 + F 2 N2F4N2F g 4.00 g 4.00 g NH 3 1 mole NH g NH g F 2 1 mole F g F 2 = moles NH 3 = moles F 2 2 ??? g moles NH 3 5 moles F 2 2 moles NH 3 =.5875 moles F 2 needed + HF 6 There is NOT ENOUGH fluorine

21 Use the L R (F 2 in this problem) to determine amounts of products… moles F 2 1 mole N 2 F 4 5 mole F g N 2 F 4 1 mole N 2 F 4 = g N 2 F 4 produced NH 3 + F 2 N2F4N2F g 4.00 g 2 ??? g + HF 6

22 Lets say you carry out the reaction described above in the lab and you only obtain 4.80 g N 2 F 4. What happened???? The quantities we calculate in mass-mass (Stoichiometry) problems are THEORETICAL AMOUNTS; that is, the maximum possible yield. In reality, the actual amount of products is much less due to human error, mechanical error, and possible side reactions. By comparing the actual versus theoretical, scientists calculate the percentage yield.

23 4.80 g N 2 F 4 actual x 100 % g N 2 F 4 theoretical From the last problem = 62.6% N 2 F 4 yielded


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