Stoichiometry

1. 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.

2. 4 4 NH3(g) + 5 O2(g) NO(g) + 6 H2O(g) 4molecules 5molecules 4molecules 6molecules 4moles 5moles 4moles 6 moles 68.16 g 160.00 g 120.04 g 108.12 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 = 6.022 x 1023 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 directlyproportional to the number of particles of the number of moles. g-formula mass = 1 mole = 6.022 x 1023 particles = 22.4 dm3 at STP* *Standard Temperature and Pressure 0C and 1.0 atm

4. 4 4 NH3(g) + 5 O2(g) NO(g) + 6 H2O(g) 4liters 5liters 4liters 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.

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

6. THE TYPICAL FORMAT: Step#3 from equation Step#4 Step#2

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: • Write the balanced equation. • Find the number of moles of the given substance. • Inspect and use the balanced equation to determine the ratio of moles of given substance to moles of “wanted” substance. • Convert the amount of “wanted” substance into the desired units. 2 2 • H2SO4(aq) • + NaOH(aq) • Na2SO4(aq) • + HOH(l) • + H2O(l) • 15 g • ??? g • = 6.8 g H2O

8. Example:How many moles of NaOH are required to neutralize 4.2 moles of H2SO4? • TO SOLVE: • Write the balanced equation. • Find the number of moles of the given substance. • Inspect and use the balanced equation to determine the ratio of moles of given substance to moles of “wanted” substance. • Convert the amount of “wanted” substance into the desired units. 2 2 • H2SO4(aq) • + NaOH(aq) • Na2SO4(aq) • + H2O(l) • 4.2 moles • ??? moles • = 8.4 moles NaOH

9. Example:How many grams of Na2SO4 are produced from 4.2 moles of H2SO4? • TO SOLVE: • Write the balanced equation. • Find the number of moles of the given substance. • Inspect and use the balanced equation to determine the ratio of moles of given substance to moles of “wanted” substance. • Convert the amount of “wanted” substance into the desired units. 2 2 • H2SO4(aq) • + NaOH(aq) • Na2SO4(aq) • + H2O(l) • ??? g • 4.2 moles • = 6.0 x 102 g Na2SO4 • = 596.568 g Na2SO4

10. Limiting Reagents So far you have worked stoichiometry problems in which the given quantity of a reactant is consumedcompletely 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 notall completely react. Lets look at the following hypothetical reaction: • C • A + B • 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 limitingreagent or reactant (LR). The LR allows you to calculate (just like you learned previously) the amount of productformed 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: Write the balanced equation for the reaction. Calculate the number of moles of each reactant. Convert one of the reactants into the other. This tells you how much of each you need to complete the reaction. Compare the moles calculated in step #2 to the amounts in #3. The one you have less of is the limiting reagent. 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 N2 and 10. g H2 are mixed, what is the limiting reagent? Write the balanced equation for the reaction. Calculate the number of moles of each reactant. 2 3 • N2(g) • + H2(g) • NH3(g) • 10. g • 20. g • = 5.0 moles H2(g) • = 0.71 moles N2(g)

15. Convert one of the reactants into the other. This tells you how much of each you need to complete the reaction. • Compare the given amounts to the answer in #3. The one you have less of is the limiting reagent. • You have 5 moles of H2…..more than enough… • N2 is the limiting reagent. There is an excess of H2, After the reaction has completed there will be no N2 left, yet there will still be extra H2. • = 2.1 moles H2 needed • = .71 N2 given and 2.1 moles need to react with it

16. EXAMPLE:How much NH3 is produced? • Use the L R (N2 in this problem) to determine amounts of products….. • From the last example, we know 3 2 • N2(g) • + H2(g) • NH3(g) • = 24 g NH3

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 3 2 • N2(g) • + H2(g) • NH3(g) • = 4.32548179871 g H2 reacted • = 6 g H2 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? 3 2 • Al(s) • + S(g) • Al2S3(s) • 9.00 g • 8.00 g • ??? g • = 0.249 moles S(s) • = 0.334 moles Al(S) • = 0.166 moles Al needed • There is more than enough aluminum Sulfur is the limiting reactant

19. Use the L R (S in this problem) to determine amounts of products….. 3 2 • Al(s) • + S(g) • Al2S3(s) • 9.00 g • 8.00 g • ??? g • = 12.5 g Al2S3 produced

20. EXAMPLE:How many grams of N2F4 can theoretically be obtained from 4.00 g of NH3 and 14.0 g of F2 according to: 5 2 6 • NH3 • + F2 • N2F4 • + HF • 4.00 g • 14.00 g • ??? g • = 0.3684 moles F2 • = 0.235 moles NH3 • = .5875 moles F2 needed • There is NOT ENOUGH fluorine Fluorine is the limiting reactant

21. Use the L R (F2 in this problem) to determine amounts of products….. 5 2 6 • NH3 • + F2 • N2F4 • + HF • 4.00 g • 14.00 g • ??? g • = 7.665 g N2F4 produced

22. Lets say you carry out the reaction described above in the lab and you only obtain 4.80 g N2F4. What happened???? The quantities we calculate in mass-mass (Stoichiometry) problems are THEORETICALAMOUNTS; that is, the maximum possible yield. In reality, the actual amount of products is muchless due to human error, mechanical error, and possible side reactions. By comparing the actual versus theoretical, scientists calculate the percentageyield.  NOTE: % Error + % Yield = 100%

23. From the last problem • = 62.6% N2F4 yielded