Stoichiometry (2)

1. Stoichiometry (2) Calculations from balanced equations

2. molecules 1 molecule C3H8 + 5 molecules O2 3 molecules CO2 + 4 molecules H2O amount (mol) 1 mol C3H8 + 5 mol O2 3 mol CO2 + 4 mol H2O mass (amu) 44.09 amu C3H8 + 160.00 amu O2 132.03 amu CO2 + 72.06 amu H2O mass (g) 44.09 g C3H8 + 160.00 g O2 132.03 g CO2 + 72.06 g H2O total mass (g) 204.09 g 204.09 g Table 3.5 Information Contained in a Balanced Equation Viewed in Terms of Reactants C3H8(g) + 5O2(g) Products 3CO2(g) + 4H2O(g)

3. Reactions in Sequence Roasting is the first step in extracting copper from chalcocite. In the next step, copper (I) oxide reacts with powdered carbon to yield copper metal and carbon monoxide gas. Determine the mass of copper which can be obtained from each gram of the original copper (I) sulfide in the ore.

4. First write balanced equations for both steps: • 2Cu2S (s) + 3O2 (g) → 2Cu2O (s) + 2SO2 (g) • 2Cu2O (s) + 2C (s) → 4 Cu (s) + 2CO (g) • Then combine: • 2Cu2S (s) + 3O2 (g) + 2 C (s) → 4 Cu (s) + 2 SO2 (g) + 2CO (g) • Now calculate, using the molar ratio from the new equation: • M of Cu2S = 159.16; Cu = 63.55 • mass of Cu = 0.799 g

5. SO2 (g) is formed in many industrial processes. This gas reacts in air with O2 to form SO3 (g). The sulfur trioxide further reacts with water to form sulfuric acid which falls as rain or snow. • Calculate the mass of sulfuric acid which can be formed from 15.0 g of SO2 released into the atmosphere.

6. Limiting Reagents • so far we have assumed reactants to be present in correct proportions • a reaction may be limited by the quantity of a specific reagent

7. An ice cream sundae analogy for limiting reactions. Figure 3.10

8. Ice-cream sundae requires: • 2 scoops ice-cream (12 oz) • 1 cherry • 50 mL syrup • You want to make 6 sundaes • You have: • 48 oz ice-cream • 6 cherries • 100 mL syrup

9. 300 oz ice-cream will make 25 sundaes • 30 cherries will make 30 sundaes • 1 L of syrup will make 20 sundaes • The syrup is limiting • The ice-cream and cherries are in excess

10. PROBLEM: Nuclear engineers use chlorine trifluoride in the processing of uranium fuel for power plants. This extremely reactive substance is formed as a gas in special metal containers by the reaction of elemental chlorine and fluorine. Sample Problem 3.10 Using Molecular Depictions to Solve a Limiting-Reactant Problem (a) Suppose the box shown at left represents a container of the reactant mixture before the reaction occurs (with chlorine colored green). Name the limiting reactant, and draw the container contents after the reaction is complete. (b) When the reaction is run again with 0.750 mol of Cl2 and 3.00 mol of F2, what mass of chlorine trifluoride will be prepared? PLAN: Write a balanced chemical equation. Compare the number of molecules you have to the number needed for the products. Determine the reactant that is in excess. The other reactant is the limiting reactant.

11. Cl2(g) + 3F2(g) 2ClF3(g) Cl2 F2 3.00 mol F2 4 = 0.750 mol Cl2 1 2 mol ClF3 92.5 g ClF3 = 139 g ClF3 1 mol Cl 1 mol ClF3 Using Molecular Depictions to Solve a Limiting-Reactant Problem Sample Problem 3.10 continued SOLUTION: (a) You need a ratio of 2 Cl and 6 F for the reaction. You have 6 Cl and 12 F. 6 Cl would require 18 F. 12 F need only 4 Cl (2 Cl2 molecules). There isn’t enough F, therefore it must be the limiting reactant. You will make 4 ClF2 molecules (4 Cl, 12 F) and have 2 Cl2 molecules left over. (b) We know the molar ratio of F2/Cl2 should be 3/1. Since we find that the ratio is 4/1, that means F2 is in excess and Cl2 is the limiting reactant. 0.750 mol Cl2

12. A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine (N2H4) and dinitrogen tetroxide, which ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00 x 102 g of N2H4 and 2.00 x 102 g of N2O4 are mixed? • 2 N2H4 + N2O4→ 3 N2 + 4H2O • N2 from N2H4 = 4.68 mol • N2 from N2O4 = 6.51 mol • thus N2H4 is limiting; mass of N2 = 131 g

13. Sample Problem 3.11 Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant PROBLEM: A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine(N2H4) and dinitrogen tetraoxide(N2O4), which ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00x102 g of N2H4 and 2.00x102 g of N2O4 are mixed? PLAN: We always start with a balanced chemical equation and find the number of mols of reactants and products which have been given. In this case one of the reactants is in molar excess and the other will limit the extent of the reaction. mass of N2H4 mass of N2O4 limiting mol N2 divide by M multiply by M mol of N2H4 mol of N2O4 g N2 molar ratio mol of N2 mol of N2

14. N2H4(l) + N2O4(l) N2(g) + H2O(l) mol N2H4 32.05g N2H4 3 mol N2 28.02g N2 2mol N2H4 mol N2 mol N2O4 92.02g N2O4 3 mol N2 mol N2O4 Sample Problem 3.11 Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant continued SOLUTION: 2 3 4 1.00x102g N2H4 = 3.12mol N2H4 N2H4 is the limiting reactant because it produces less product, N2, than does N2O4. 3.12mol N2H4 = 4.68mol N2 4.68mol N2 = 131g N2 2.00x102g N2O4 = 2.17mol N2O4 2.17mol N2O4 = 6.51mol N2

15. Consider the following reaction:2 Na3PO4(aq) + 3 Ba(NO3)2(aq) → Ba3(PO4)2(s) + 6 NaNO3(aq)Suppose a solution containing 3.50 g of Na3PO4 is mixed with a solution containing 6.40 g of Ba(NO3)2. How many grams of Ba3(PO4)2 can be formed? • Na3PO4 = 164; moles = 0.0213; product = 0.01065 mol • 3 Ba(NO3)2 = 261; moles = 0.0245; pr = 0.00817 mol • thus Ba(NO3)2 is limiting; • product = 4.92 g

16. A strip of zinc metal weighing 2.00 g is placed in an aqueous solution containing 2.50 g of silver nitrate, causing the following reaction to occur:Zn (s) + 2AgNO3 (aq) → Zn(NO3)2 (aq) + 2Ag (s)Determine which reactant is limiting, and calculate the mass of solid Ag which can be formed. • AgNO3 = 169.88; moles = 0.0147; pr = 0.0147 mol • Zn = 65.39; moles = 0.0306; pr = 0.0612 mol • thus AgNO3 is limiting; mass of product = 1.59 g

17. How many grams of s solid aluminum sulfide can be prepared by the reaction of 10.0 g of aluminum and 15.0 g of sulfur? How much of the non-limiting reagent is in excess? • 2Al + 3S → Al2S3 • product from Al = 0.185 mol • product from S = 0.156 mol • thus S is limiting; product from S = 23.4 g (M = 149.99) • use 0.312 mol of Al; • thus 0.371 – 0.312 = 0.059 mol left = 1.6 g

18. Reaction Yield • quantities calculated from balanced equation are theoretical • in practice, side reactions occur or unreacted reactants remain • yield is usually lower than expected

19. determined by weighing calculated from equation

20. Silicon carbide (SiC) is an important ceramic material that is made by allowing sand (SiO2) to react with powdered carbon at high temperature. Carbon monoxide is also formed. When 100.0 kg of sand are processed, 51.4 kg of SiC are recovered. Determine the percentage yield in this process. • SiO2 (s) + 3 C (s) → SiC (s) + 2 CO (g) • SiC = 41.0 g; SiO2 = 60.09 g • theoretical yield = 66.73 kg • % yield = 77.0%

21. Marble (calcium carbonate) reacts with hydrochloric acid solution to form calcium chloride solution, water, and carbon dioxide. What is the % yield of carbon dioxide if 3.65 g of the gas is collected when 10.0 g of marble reacts? • CaCO3 (s) + 2 HCl (aq) → CaCl2 (aq) + H2O (l) + CO2 (g) • CaCO3 = 100.091; CO2 = 44.011 • theoretical yield = 4.40 g • % yield = 83.0%