Stoichiometry STOICHIOMETRY IS THE PART OF CHEMISTRY THAT

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Stoichiometry STOICHIOMETRY IS THE PART OF CHEMISTRY THAT DEALS WITH THE AMOUNTS OF SUBSTANCES

Stoichiometry 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.

4 NH 3(g) + 5 O 2(g) 4 molecules 5 molecules 4 NO(g) +

4 NH 3(g) + 5 O 2(g) 4 molecules 5 molecules 4 NO(g) + 6 H 2 O(g) 4 molecules 6 molecules 4 moles 5 moles 4 moles 68. 16 g 160. 00 g 120. 04 g 108. 12 g Recall the definition of a mole: g-formula mass = 1 mole = 6. 022 x 1023 particles 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.

In 1811 Avogadro postulated his own gas law. He said that, “Equal volumes of

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 = 6. 022 x 1023 particles = 22. 4 dm 3 at STP* *Standard Temperature and Pressure 0 C and 1. 0 atm

4 NH 3(g) + 5 O 2(g) 4 liters 5 liters 4 NO(g) +

4 NH 3(g) + 5 O 2(g) 4 liters 5 liters 4 NO(g) + 6 H 2 O(g) 4 liters 6 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.

TO SOLVE: 1. Write the balanced equation. 2. Find the number of moles of

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.

THE TYPICAL FORMAT: (given) grams 1 mole (given) Step#2 g-formula mass (given) moles of

THE TYPICAL FORMAT: (given) grams 1 mole (given) Step#2 g-formula mass (given) moles of wanted Step#3 from equation moles of given g-formula mass (wanted) Step#4 1 mole (wanted)

Example: Sulfuric acid is neutralized by the addition of sodium hydroxide. How many grams

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 H 2 SO 4(aq) + Na. OH(aq) 15 g 2 Na 2 SO 4(aq) + HOH + H 2 O(l) ? ? ? g 2. Find the number of moles of the given substance. 15 g Na. OH 1 mole Na. OH 2 mole H 2 O 18. 02 g H 2 O 40. 00 g Na. OH 2 mole Na. OH 1 mole H 2 O = 6. 8 g H 2 O 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.

Example: How many moles of Na. OH are required to neutralize 4. 2 moles

Example: How many moles of Na. OH are required to neutralize 4. 2 moles of H 2 SO 4? TO SOLVE: 1. Write the balanced equation. 2 H 2 SO 4(aq) + Na. OH(aq) 4. 2 moles ? ? ? moles 2 Na 2 SO 4(aq) + H 2 O(l) 2. Find the number of moles of the given substance. 4. 2 moles H 2 SO 4 2 moles Na. OH 1 mole H 2 SO 4 = 8. 4 moles Na. OH 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.

Example: How many grams of Na 2 SO 4 are produced from 4. 2

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 H 2 SO 4(aq) + Na. OH(aq) 4. 2 moles 2 Na 2 SO 4(aq) + H 2 O(l) ? ? ? g 2. Find the number of moles of the given substance. 4. 2 moles H 2 SO 4 1 mole Na 2 SO 4 142. 04 g. Na 2 SO 4 1 mole H 2 SO 4 1 mole Na 2 SO 4 = 596. 568 g Na 2 SO 4 = 6. 0 x 102 g Na 2 SO 4 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.

Limiting Reagents So far you have worked stoichiometry problems in which the given quantity

Limiting Reagents 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.

The reagent that is consumed 100% is referred to as the limiting reagent or

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.

Limiting reactant analogies…. For example, bicycles require one frame and two wheels. If you

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.

TO SOLVE LIMITING REACTANT (REAGENT) PROBLEMS: 1. Write the balanced equation for the reaction.

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.

EXAMPLE: The Haber Process is used to convert nitrogen and hydrogen gases to ammonia.

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? 1. Write the balanced equation for the reaction. N 2(g) 20. g 3 + H 2(g) 10. g 2 NH 3(g) 2. Calculate the number of moles of each reactant. 20. g N 2 1 mole N 2 28. 02 g N 2 = 0. 71 moles N 2(g) 10. g H 2 1 mole H 2 2. 02 g H 2 = 5. 0 moles H 2(g)

3. Convert one of the reactants into the other. This tells you how much

3. Convert one of the reactants into the other. This tells you how much of each you need to complete the reaction. . 713775874375 moles N 2 3 mole H 2 1 mole N 2 = 2. 1 moles H 2 needed 4. Compare the given amounts to the answer in #3. The one you have less of is the limiting reagent. =. 71 N 2 given and 2. 1 moles need to react with it • 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.

EXAMPLE: How much NH 3 is produced? Use the L R (N 2 in

EXAMPLE: How much NH 3 is produced? Use the L R (N 2 in this problem) to determine amounts of products…. . N 2(g) 3 2 NH + H 2(g) 3(g) From the last example, we know. 713775874375 moles N 2 2 mole NH 3 17. 04 g NH 3 1 mole N 2 1 mole NH 3 = 24 g NH 3

EXAMPLE: How much excess is there? Use the L R (N 2 in this

EXAMPLE: How much excess is there? Use the L R (N 2 in this problem) to determine how much of the other reactant (H 2 in this problem) is left over. N 2(g) 3 2 NH + H 2(g) 3(g) From the last example, we know. 713775874375 moles N 2 3 mole H 2 2. 02 g. H 2 1 mole N 2 1 mole H 2 = 4. 32548179871 g H 2 reacted 10. g H 2 to start - 4. 325481 g H 2 reacted 5. 674518 g in excess = 6 g H 2 left over

EXAMPLE: How many grams of aluminum sulfide can form from the reaction of 9.

EXAMPLE: How many grams of aluminum sulfide can form from the reaction of 9. 00 g of aluminum with 8. 00 g of sulfur? 2 Al (s) 9. 00 g 1 mole Al 9. 00 g Al Al 2 S 3(s) 8. 00 g ? ? ? g 8. 00 g S 26. 98 g Al 1 mole S 32. 07 g S = 0. 334 moles Al(S) 0. 249 moles S 3 + S(g) 2 moles Al 3 moles S = 0. 249 moles S(s) = 0. 166 moles Al needed There is more than enough aluminum Sulfur is the limiting reactant

Use the L R (S in this problem) to determine amounts of products…. .

Use the L R (S in this problem) to determine amounts of products…. . 2 Al 3 + S(g) (s) 9. 00 g 0. 249454318678 moles S Al 2 S 3(s) 8. 00 g ? ? ? g 1 mole Al 2 S 3 150. 17 g Al 2 S 3 3 mole S 1 mole Al 2 S 3 = 12. 5 g Al 2 S 3 produced

EXAMPLE: How many grams of N 2 F 4 can theoretically be obtained from

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: 2 NH 3 4. 00 g NH 3 5 + F 2 N 2 F 4 14. 00 g ? ? ? g 1 mole NH 3 14. 00 g F 2 17. 04 g NH 3 1 mole F 2 38. 00 g F 2 = 0. 3684 moles F 2 = 0. 235 moles NH 3 6 + HF 5 moles F 2 2 moles NH 3 =. 5875 moles F 2 needed There is NOT ENOUGH fluorine Fluorine is the limiting reactant

Use the L R (F 2 in this problem) to determine amounts of products….

Use the L R (F 2 in this problem) to determine amounts of products…. . 2 NH 3 4. 00 g 0. 368421052632 moles F 2 5 + F 2 14. 00 g N 2 F 4 ? ? ? g 1 mole N 2 F 4 104. 02 g N 2 F 4 5 mole F 2 1 mole N 2 F 4 = 7. 665 g N 2 F 4 produced 6 + HF

Lets say you carry out the reaction described above in the lab and you

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. Actual % Yield = Theoretical NOTE: % Error + % Yield = 100%

% Yield = Actual Theoretical X 100% From the last problem 4. 80 g

% Yield = Actual Theoretical X 100% From the last problem 4. 80 g N 2 F 4 actual 7. 665 g N 2 F 4 theoretical x 100 % = 62. 6% N 2 F 4 yielded