CHEMICAL REACTIONS CHAPTER 4 Reactants: Zn + I2 Product: ZnI2

Chapter 4 Outline Chemical Equations Stoichiometry Limiting Reactants Chemical Analysis

Chemical Equations 4 Al(s) + 3 O2(g) 2Al2O3(s) Depict the kind of reactants and products and their relative amounts in a reaction 4 Al(s) + 3 O2(g) 2Al2O3(s) The numbers in the front are called stoichiometric coefficients The letters (s), (g), and (l) are the physical states of compounds.

Chemical Equations 4 Al(s) + 3 O2(g) 2 Al2O3(s) This equation means 4 Al atoms + 3 O2 molecules 2 molecules of Al2O3 4 moles of Al + 3 moles of O2 2 moles of Al2O3

Chemical Equations Because the same atoms are present in a reaction at the beginning and at the end, the amount of matter in a system does not change. The Law of the Conservation of Matter

Chemical Equations Lavoisier, 1788 Because of the principle of the conservation of matter, an equation must be balanced. It must have the same number of atoms of the same kind on both sides. Lavoisier, 1788

Balancing Equations

Balancing Equations

Balancing Equations ___C3H8 (g) + ___ O2 (g) ----> ___CO2 (g) + ___ H2O (g) C3H8 (g) + 5 O2 (g) ----> 3 CO2 (g) + 4 H2O (g)

Balancing Equations ___B4H10 (g) + ___ O2 (g) ---> ___ B2O3 (g) + ___ H2O (g) 2 B4H10 (g) + 11 O2 (g) ---> 4 B2O3 (g) + 10 H2O (g)

11 STOICHIOMETRY - the study of the quantitative aspects of chemical reactions.

PROBLEM: If 454 g of NH4NO3 decomposes, how much H2O and N2O are formed? What is the theoretical yield of products? STEP 1 Write the balanced chemical equation NH4NO3 N2O + 2 H2O

Convert reactant mass to moles 454 g of NH4NO3 --> N2O + 2 H2O STEP 2 Convert reactant mass to moles (454 g) --> moles 454 g • 1 mol 80.04 g = 5.68 mol NH 4 NO 3

2 mol H O produced 1 mol NH NO used 4 3 STEP 3 454 g of NH4NO3 --> N2O + 2 H2O STEP 3 Convert moles reactant --> moles product. Relate moles NH4NO3 to moles product. 1 mol NH4NO3 --> 2 mol H2O Express this relation as the STOICHIOMETRIC FACTOR 2 mol H O produced 1 mol NH 4 NO 3 used

Convert moles reactant (5.68 mol) moles product 454 g of NH4NO3 --> N2O + 2 H2O STEP 3 Convert moles reactant (5.68 mol) moles product 5.68 mol NH 4 NO 3 • 2 mol H O produced 1 mol NH used = 11.4 mol H2O produced

STEP 4 454 g of NH4NO3 --> N2O + 2 H2O THEORETICAL YIELD Convert moles product (11.4 mol) to mass product. This is called the THEORETICAL YIELD This is the “Expected” # of moles.

STEP 4 454 g of NH4NO3 --> N2O + 2 H2O Convert moles prod. (11.4 mol) to mass prod. 18 . 02 g 11.4 mol H O • = 204 g H O 2 2 1 mol This is the “Expected” Mass! ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS! Repeat to find the grams of N2O formed.

STEP 5 454 g of NH4NO3 --> N2O + 2 H2O How much N2O is formed? Total mass of reactants=total mass of products 454 g NH4NO3 = ___ g N2O + 204 g H2O mass of N2O = 250. g This is an alternate method.

STEP 6 454 g of NH4NO3 --> N2O + 2 H2O Calculate the percent yield. If you isolated only 131 g of N2O, what is the percent yield? This compares the theoretical (250. g) and actual (131 g) yields.

Calculate the percent yield. 454 g of NH4NO3 --> N2O + 2 H2O STEP 6 Calculate the percent yield. % yield = actual yield theoretical yield • 100% % yield = 131 g 250. g • 100% 52.4%

Stoichiometry Calculations General Plan For Stoichiometry Calculations Mass reactant Stoichiometric factor Moles product

2 H2O2(liq) ---> 2 H2O(g) + O2(g) Reaction is catalyzed by MnO2 PROBLEM: Using 5.00 g of H2O2, what mass of O2 and of H2O can be obtained? 2 H2O2(liq) ---> 2 H2O(g) + O2(g) Reaction is catalyzed by MnO2 Step 1: moles of H2O2 Step 2: use STOICHIOMETRIC FACTOR to calculate moles of O2 Step 3: mass of O2 Repeat for H2O.

Reactions Involving a LIMITING REACTANT In a given reaction, there is not enough of one reagent to use up the other reagent completely. The reagent in short supply LIMITS the quantity of product that can be formed.

LIMITING REACTANTS 2 NO(g) + O2 (g) 2 NO2(g) NO Products 2 NO(g) + O2 (g) 2 NO2(g) Limiting reactant = ___________ Excess reactant = ____________ NO O2

LIMITING REACTANTS

LIMITING REACTANTS React solid Zn with 0.100 mol HCl (aq) Zn + 2 HCl ---> ZnCl2 + H2 Rxn 1 Rxn 2 Rxn 3 mass Zn (g) 7.00 3.27 1.31 mol Zn 0.107 0.050 0.020 mol HCl 0.100 0.100 0.100 mol HCl/mol Zn 0.93 2.00 5.00

PROBLEM: Mix 5. 40 g of Al with 8. 10 g of Cl2 PROBLEM: Mix 5.40 g of Al with 8.10 g of Cl2. How many grams of Al2Cl6 can form?

Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.

mol Cl mol Al = 3 2 Al + 3 Cl2 ---> Al2Cl6 2 Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio. 2 Al + 3 Cl2 ---> Al2Cl6 Reactants must be in the mole ratio mol Cl 2 mol Al = 3

Deciding on the Limiting Reactant 2 Al + 3 Cl2 Al2Cl6 mol Cl 2 mol Al > 3 If then there is not enough Al to use up all the Cl2, and the limiting reagent is Al.

Deciding on the Limiting Reactant 2 Al + 3 Cl2 ---> Al2Cl6 mol Cl 2 mol Al < 3 If then there is not enough Cl2 to use up all the Al, and the limiting reagent is Cl2

Step 2 of LR problem: Calculate moles of each reactant We have 5.40 g of Al and 8.10 g of Cl2 5 . 40 g Al • 1 mol 27.0 g = 0.200 mol Al 1 mol 8 .10 g Cl • = 0.114 mol Cl 2 2 70.9 g

Find mole ratio of reactants Cl 2 mol Al = 0.114 mol 0.200 mol 0.57 This should be 3/2 or 1.5/1 if reactants are present in the exact stoichiometric ratio. Limiting reagent is Cl2

Base all calculations on Cl2 Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form? Limiting reactant = Cl2 Base all calculations on Cl2 moles Cl2 Al2Cl6 grams

CALCULATIONS: calculate mass of Al2Cl6 expected. Step 1: Calculate moles of Al2Cl6 expected based on LR. . 114 mol Cl 2 • 1 Al 6 3 mol Cl = 0.0380 mol Al Step 2: Calculate mass of Al2Cl6 expected based on LR. . 0380 mol Al 2 Cl 6 • 66.4 g Al = 10.1 g Al

How much of which reactant will remain when reaction is complete? Cl2 was the limiting reactant. Therefore, Al was present in excess. But how much? First find how much Al was required. Then find how much Al is in excess.

Calculating Excess Al 2 Al + 3 Cl2 products 0.114 mol = LR 0.200 mol

Calculating Excess Al 2 Al + 3 Cl2 0.114 mol = LR products 0.114 mol = LR 0.200 mol 2 mol Al .114 mol Cl • = 0.0760 mol Al req' d 2 3 mol Cl 2 Excess Al = Al available - Al required = 0.200 mol - 0.0760 mol = 0.124 mol Al in excess

Nitrogen and iodine react to form nitrogen tri iodide. If 50 Nitrogen and iodine react to form nitrogen tri iodide. If 50.0 g of nitrogen is mixed with 350.0 g iodine, calculate the number of grams of product formed and the grams of reactant remaining. N2 + 3 I2 2 NI3 28.0 g/mol 253.8 g/mol 394.7 g/mol 1.79 mole 1.38 mole L.R. (0.460 S.U.) 50.0g -12.9g 0g left 363 g 37.1g left 350.0g + 50.0g = 400.0g = 363g + 37.1g

Using Stoichiometry to Determine a Formula Burn 0.115 g of a hydrocarbon, CxHy, and produce 0.379 g of CO2 and 0.1035 g of H2O. What is the empirical formula of CxHy? CxHy + oxygen 0.379 g CO2 + 0.1035 g H2O 40

Chemical Molecular Analysis in the lab

Using Stoichiometry to Determine a Formula CxHy + some oxygen 0.379 g CO2 + 0.1035 g H2O First, recognize that all C in CO2 and all H in H2O is from CxHy. 1. Calculate moles of C in CO2 8.61 x 10-3 mol C 2. Calculate moles of H in H2O 1.149 x 10 -2 mol H

Using Stoichiometry to Determine a Formula CxHy + some oxygen 0.379 g CO2 + 0.1035 g H2O Now find ratio of mol H/mol C to find values of x and y in CxHy. 1.149 x 10 -2 mol H/ 8.61 x 10-3 mol C = 1.33 mol H / 1.00 mol C = 4 mol H / 3 mol C Empirical formula = C3H4

Chemical Analysis Combustion-Determine a Formula CxHy + some oxygen 0.379 g CO2 + 0.1035 g H2O Recognize that all C in CO2 and all H in H2O is from CxHy. 1. Calculate moles of C in CO2 0.379 g CO2 mole CO2 mole C 44.0 g CO2 mole CO2 = 0.00861 mole C

Chemical Analysis Combustion-Determine a Formula CxHy+some oxygen 0.379gCO2+0.1035 g H2O 2. Calculate moles of H in H2O 0.1035 g H2O mole H2O 2 mole H 18.0 g H2O mole H2O = 0.0115 mole H

Chemical Analysis Combustion-Determine a Formula CxHy + some oxygen 0.379 g CO2 + 0.1035 g H2O Now find ratio of mol H/mol C to find values of x and y in CxHy. 0.0115 mol H/ 0.00861 mol C = 1.34 mol H / 1.00 mol C = 4.02 mol H / 3.00 mol C Empirical formula = C3H4

Sample Problems 1) A 0.537 g sample of an unknown compound containing only carbon, hydrogen, and oxygen was burned to produced 1.030 g of CO2 and 0.632 g of H2O. Determine the empirical formula. Given that the molecular weight is approximately 90 g/mole, determine the molecular formula. 1.030 g CO2 mole CO2 mole C 12.0 g C 44.0g CO2 mole CO2 mole C = 0.281 g C 0.632 g H2O mole H2O 2 mole H 1.0 g H 18.0 g H2O mole H2O mole H = 0.070 g H

Sample Problems 0.537 g C, H, O - 0.281 g C - 0.070 g H 0.186 g O 1) A 0.537 g sample of an unknown compound containing only carbon, hydrogen, and oxygen was burned to produced 1.030 g of CO2 and 0.632 g of H2O. Determine the empirical formula. Given that the molecular weight is approximately 90 g/mole, determine the molecular formula. 0.537 g C, H, O - 0.281 g C - 0.070 g H 0.186 g O

Sample Problems C H O .281 g 1 mole 12.0 g .070 g 1 mole 1.0 g .186 g 0.0234 0.070 0.0116 0.0116 0.0116 0.0116 2.02 6.0 1.00 Empirical formula C2H6O

Sample Problems Empirical formula C2H6O 90 46 = 2 Molecular formula C4H12O2 Alternate method. Use grams of cmpd. and its molar mass to determine moles of cmpd. Divide moles of C and H by these moles to find the subscripts for C and H. The subscript for O can be determinded by difference.

Sample Problems 2) A 0.1247 g sample of ascorbic acid, vitamin C, was burned to produce 0.1869 g of CO2 and 0.0510 g H2O. Ascorbic acid contains only carbon, hydrogen, and oxygen. Determine the empirical formula. Given that the molecular weight is approximately 180 g/mole, determine the molecular formula. .1869 g CO2 mole CO2 mole C 12.0 g C 44.0 g CO2 mole CO2 mole C = 0.0510 g C .0510 g H2O mole H2O 2 mole H 1.0 g H 18.0 g H2O mole H2O mole H = 0.0057 g H

Sample Problems 0.1247 g C, H, O - 0.0510 g C - 0.0057 g H 0.0680 g O 2) A 0.1247 g sample of ascorbic acid, vitamin C, was burned to produce 0.1869 g of CO2 and 0.0510 g H2O. Ascorbic acid contains only carbon, hydrogen, and oxygen. Determine the empirical formula. Given that the molecular weight is approximately 180 g/mole, determine the molecular formula. 0.1247 g C, H, O - 0.0510 g C - 0.0057 g H 0.0680 g O

Sample Problems C H O .0510g 1 mole 12.0 g .0057g 1 mole 1.0 g .0680g .00425 .0057 .00425 .00425 .00425 .00425 1.00 1.3 1.00 3.00 3.9 3.00 Empirical formula C3H4O3

Sample Problems Empirical formula C3H4O3 180 = 2 88 Molecular formula C6H8O6

Chemical Analysis Mixtures-Determine a Percent 1. The amount of calcium present in milk can be determined by adding oxalate ion, C2O42-(in the form of its water-soluble sodium salt, Na2C2O4); the insoluble compound calcium oxalate is precipitated. Suppose you take a 75.0 g sample of milk and isolate 0.288 g of calcium oxalate from it. What is the weight percentage of calcium in the milk? MILK (Ca2+) Na2C2O4 CaC2O4

Chemical Analysis Mixtures-Determine a Percent MILK (Ca2+) Na2C2O4 CaC2O4 75.0 g 0.288 g ? % Ca % Ca = g Ca 75.0 g milk X 100

Chemical Analysis Mixtures-Determine a Percent MILK (Ca2+) Na2C2O4 CaC2O4 75.0 g 0.288 g ? % Ca .288 g CaC2O4 mole CaC2O4 mole Ca 40.1 g Ca 128.1 g CaC2O4 mole CaC2O4 mole Ca = 0.0902 g Ca % Ca = 0.0902 g Ca 75.0 g milk X 100 = 0.120%

Chemical Analysis Mixtures-Determine a Percent 2. A 4.22 g sample of calcium chloride and sodium chloride was dissolved in water, and the solution was treated with sodium carbonate to precipitate the calcium as calcium carbonate. After isolating the solid calcium carbonate, it was heated to drive off the carbon dioxide and form 0.959 g of calcium oxide. What is the weight percent of calcium chloride in the original 4.22 g sample?

Chemical Analysis Mixtures-Determine a Percent CaCl2/NaCl Na2CO3 CaCO3 4.22 g Heat % CaCl2 = g CaCl2 4.22 g sample X 100 CaO 0.959 g

Chemical Analysis Mixtures-Determine a Percent 0.959 g CaO mole CaO mole CaCl2 111.1 g CaCl2 56.1 g CaO mole CaO 4 mole CaCl2 = 1.90 g CaCl2 % CaCl2 = 1.90 g CaCl2 4.22 g sample X 100 = 45.0%

Practice Problems 1. Balance the following equations: CS2 + Cl2 --> CCl4 + S2Cl2 N2 + O2 --> NO C8H18 + O2 --> CO2 + H2O 2. Write the formula equation for each of the following: sodium + water --> sodium hydroxide + hydrogen magnesium + oxygen --> magnesium oxide aluminum + hydrochloric acid --> aluminum chloride + hydrogen

Practice Problems 2. (continue) aluminum + hydrochloric acid --> aluminum chloride + hydrogen sodium chlorate --> sodium chloride + oxygen mercury(II) sulfate + ammonium sulfide --> mercury(II) sulfide + ammonium sulfate iron + cupric sulfate --> iron(III) sulfate + copper

Practice Problems For problems 3-6 3 H2 + N2 g 2 NH3 3. How many moles of H2 are required to react 4.2 moles of N2? 4. How many moles of H2 are required to react 74 grams of N2? 5. How many grams of NH3 would be produced from 45 g of H2? 6. How many moles of NH3 would be produced from the reaction of 18.5 g H2 and 95 g of N2?

Practice Problems 7. Phosphine, PH3, is formed when calcium phosphide is added to water. How many grams of phosphine can be obtained from 205 g of calcium phosphide? 8. How many grams of iron will be required to release all of the antimony from 10.0 g antimony trisulfide? (Ferrous sulfide is formed) 9. If calcium oxide were prepared by heating calcium carbonate, how many grams of the carbonate would be required to produce the 15.0 g of the oxide?

Practice Problems 10. How many grams of cupric sulfate are needed to completely react with 145 g of sodium chloride? How many grams of sodium sulfate could be produced by this reaction? 11. How many grams of sulfuric acid will react with 40.0 g of aluminum metal? 12. 75 grams of zinc are added to 120 grams of nitrous acid. How many grams of hydrogen gas are evolved?

Practice Problems 13. 10.0 grams of hydrogen and 75 grams of oxygen are exploded together in a reaction tube. How many grams of water are produced? What other gas is found in the tube(besides water vapor) after the reaction, and how many grams of this gas are there?

Practice Problems 14. A white powder was a mixture of NaBr and NaNO3. A sample of the powder weighing 1.341 g was dissolved in water and a solution of AgNO3 was added until the precipitation of AgBr was complete. The reaction mixture was filtered and dried and the precipitate of AgBr weighed 1.896 g. What was the percentage by weight of NaBr in the original sample?

Practice Problems 15. A 4.81 g sample of an unknown compound containing only carbon, hydrogen, and nitrogen was burned to produce 13.74 g of CO2 and 1.68 g of H2O. Determine the empirical formula.

Practice Problems Answers 1. 1,3,1,1 1,1,2 2,25,16,18 2. 2 Na + 2 HOH --> 2 NaOH + H2 2 Mg + O2 --> 2 MgO 2 Al + 6 HCl --> 2 AlCl3 + 3 H2 2 NaClO3 --> 2 NaCl + 3 O2 HgSO4 + (NH4)2S --> Hg + (NH4)2SO4 2 Fe + 3 CuSO4 --> Fe2(SO4)3 + 3 Cu 3. 13 mole H2 4. 7.9 mole H2 5. 260 g NH3 6. 6.2 mole NH3

Practice Problems Answers 7. 76.5 g PH3 8. 7.68 g Fe 9. 26.8 g CaCO3 10. 198 g CuSO4, 176 g Na2SO4 11. 218 g H2SO4 12. 2.3 g H2 13. 84 g H2O, 0.6 g H2 14. 77.48% 15. C5H3N

Writing Equations zinc + chlorine ---> zinc chloride Zn (s) + Cl2 (g) --> ZnCl2 (s) Combination, Synthesis

Writing Equations KNO3 (s) --> KNO2 (s) + O2 (g) potassium nitrate --> potassium nitrite + oxygen KNO3 (s) --> KNO2 (s) + O2 (g) 2 KNO3 (s) --> 2 KNO2 (s) + O2 (g) Decomposition

Writing Equations Single Displacement magnesium bromide + chlorine --> magnesium chloride + bromine MgBr2 (s) + Cl2 (g) --> MgCl2 (s) + Br2 (g) Single Displacement

Writing Equations Double Displacement calcium hydroxide + hydrochloric acid --> calcium chloride + water Ca(OH)2 (aq) + HCl (aq) --> CaCl2 (aq) + H2O (l) Ca(OH)2 (aq) + 2HCl (aq) --> CaCl2 (aq) + 2 H2O (l) Double Displacement

Writing Equations zinc chloride + ammonium sulfide --> zinc sulfide + ammonium chloride ZnCl2 (aq) + (NH4)2S (aq) --> ZnS (s) + NH4Cl (aq) ZnCl2 (aq) + (NH4)2S (aq) --> ZnS (s) + 2 NH4Cl (aq)

Writing Equations aluminum + cupric chloride --> copper + aluminum chloride Al (s) + CuCl2 (aq) --> Cu (s) + AlCl3 (aq) 2 Al (s) + 3 CuCl2 (aq) --> 3 Cu (s) + 2 AlCl3 (aq)

PROBLEM: How many moles of H2 are required to produce 9 moles of NH3? STEP 1 Write the balanced chemical equation 3 H2 + N2 --> 2 NH3

STEP 2 Write the given and requested information below the equation. 3 H2 + N2 --> 2 NH3 ? mole 9 mole

STEP 3 9 mole NH3 3 mole H2 2 mole NH3 = 10 mole H2 Calculate using the information. 3 H2 + N2 --> 2 NH3 ? mole 9 mole 9 mole NH3 3 mole H2 2 mole NH3 = 10 mole H2

PROBLEM: How many moles of NH3 can be produced from 10.4 moles of N2? 3 H2 + N2 --> 2 NH3 10.4 mole ? mole 10.4 mole N2 2 mole NH3 mole N2 = 20.8 mole NH3

PROBLEM: How many grams of H2 are required to produce 8 moles of NH3? 3 H2 + N2 --> 2 NH3 ? g 8 mole 8 mole NH3 3 mole H2 2.0 g H2 2 mole NH3 mole H2 = 20 g H2

PROBLEM: How many moles of NH3 can be produced from 55 grams of N2? 3 H2 + N2 --> 2 NH3 55 g ? mole 55 g N2 1 mole N2 2 mole NH3 28.0 g N2 mole N2 = 3.9 mole NH3

PROBLEM: How many grams of H2 are required to react 24 grams of N2? 3 H2 + N2 --> 2 NH3 ? g 24 g 24 g N2 mole N2 3 mole H2 2.0 g H2 28.0 g N2 mole N2 mole H2 = 5.1 g H2

PROBLEM: How many grams of N2 are required to produce 155 grams of NH3? 3 H2 + N2 --> 2 NH3 ? g 155 g 155 g NH3 mole NH3 mole N2 28.0 g N2 17.0 g NH3 2 mole NH3 mole N2 = 128 g N2

Sample Problems 1) Sulfur dioxide may be oxidized to sulfur trioxide. How many grams of sulfur dioxide could be converted by this process if 100.0 g of oxygen are available for the oxidation? 2 SO2 + O2 --> 2 SO3 ? g 100.0 g 100.0 g O2 mole O2 2 mole SO2 64.1 g SO2 32.0 g O2 mole O2 mole SO2 = 401 g SO2

Sample Problems 2) Lightning discharges in the atmosphere catalyze the conversion of nitrogen to nitrogen dioxide. How many grams of nitrogen would be required to make 25.0 g of nitrogen dioxide in this way? N2 + 2 O2 --> 2 NO2 ? g 25.0 g 25.0 g NO2 mole NO2 1 mole N2 28.0 g N2 46.0 g NO2 2 mole NO2 mole N2 = 7.61 g N2

Sample Problems 3) Ferric oxide may be reduced to pure iron with coke (pure carbon). Suppose that 150.0 g of ferric oxide is available. How many grams of carbon would be needed? 2 Fe2O3 + 3 C 4 Fe + 3 CO2 150.0 g ? g 150.0 g Fe2O3 mole Fe2O3 3 mole C 12.0 g C 159.6 g Fe2O3 2 mole Fe2O3 mole C = 16.9 g C

Sample Problems 4) Zinc metal will react with hydrochloric acid to produce hydrogen gas. If 50.0 g of zinc is to be used in the reaction, how many grams of acid would be needed to completely react with all of the zinc? How many grams of hydrogen gas would be produced? Zn + 2 HCl --> ZnCl2 + H2 50.0 g ? g ? g 50.0 g Zn mole Zn 2 mole HCl 36.5 g HCl 65.4 g Zn mole Zn mole HCl = 55.8gHCl

Sample Problems 50.0 g Zn mole Zn mole H2 2.0 g H2 4) Zinc metal will react with hydrochloric acid to produce hydrogen gas. If 50.0 g of zinc is to be used in the reaction, how many grams of acid would be needed to completely react with all of the zinc? How many grams of hydrogen gas would be produced? Zn + 2 HCl --> ZnCl2 + H2 50.0 g ? g ? g 50.0 g Zn mole Zn mole H2 2.0 g H2 65.4 g Zn mole Zn mole H2 = 1.5 g H2

Sample Problems 5) Phosphoric acid, H3PO4, is produced in the reaction between calcium phosphate and sulfuric acid. How many grams of phosphoric acid would be produced from 55 grams of calcium phosphate? What other product is formed and in what quantity? Ca3(PO4)2 + 3 H2SO4 --> 3 CaSO4 + 2 H3PO4 55 g ? g ? g 55 g Ca3(PO4)2 mole Ca3(PO4)2 3 mole CaSO4 136.2 g CaSO4 310.3 gCa3(PO4)2 mole Ca3(PO4)2 mole CaSO4 = 72 g CaSO4

Sample Problems 5) Phosphoric acid, H3PO4, is produced in the reaction between calcium phosphate and sulfuric acid. How many grams of phosphoric acid would be produced from 55 grams of calcium phosphate? What other product is formed and in what quantity? Ca3(PO4)2 + 3 H2SO4 --> 3 CaSO4 + 2 H3PO4 55 g ? g ? g 55 g Ca3(PO4)2 mole Ca3(PO4)2 2 mole H3PO4 98.0 g H3PO4 310.3 gCa3(PO4)2 mole Ca3(PO4)2 mole H3PO4 = 35 g H3PO4

Percent Yield % yield = actual yield theoretical yield • 100%

PROBLEM: If 19.3 g H2 produces 78.5 g NH3, what is the percent yield? 3 H2 + N2 --> 2 NH3 19.3 g ? g 19.3 g H2 mole H2 2 mole NH3 17.0 g NH3 2.0 g H2 3 mole H2 mole NH3 = 110 g NH3 78.5g 110 g % yield = x 100 = 71 %

PROBLEM: If the yield obtained is 75%, how many grams of NH3 would be obtained from 10.4 g of N2? 3 H2 + N2 --> 2 NH3 10.4 g ? g 10.4 g N2 mole N2 2 mole NH3 17.0 g NH3 28.0 g N2 mole N2 mole NH3 = 12.6 g NH3 12.6 g NH3 x .75 = 9.4 g NH3

PROBLEM: 75.0 grams of potassium hydroxide are permitted to react with 50.0 grams of hydrochloric acid. How many grams of potassium chloride are formed? STEP 1 Write the balanced chemical equation KOH + HCl --> KCl + HOH

STEP 2 Write the given and requested information below the equation. KOH + HCl --> KCl + HOH 75.0 g 50.0 g ?g

STEP 3 75.0 g KOH mole KOH mole KCl 74.6 g KCl Calculate the product assuming that each reactant is the limiting reagent. KOH + HCl --> KCl + HOH 75.0 g 50.0 g ?g 75.0 g KOH mole KOH mole KCl 74.6 g KCl 56.1 g KOH mole KOH mole KCl = 99.7g KCl 50.0 g HCl mole HCl mole KCl 74.6 g KCl 36.5 g HCl mole HCl mole KCl = 102 g KCl

STEP 4 75.0 g KOH can produce 99.7 g KCl Determine the limiting reactant and the actual amount of product. KOH + HCl --> KCl + HOH 75.0 g 50.0 g ?g 75.0 g KOH can produce 99.7 g KCl KOH is the limiting reactant, HCl is the excess reactant. 50.0 g HCl can produce 102 g KCl 99.7 g KCl

STEP 5 75.0 g KOH mole KOH mole HCl 36.5 g HCl Determine the amount of excess reactant by calculating the amount used and subtracting from the starting amount. KOH + HCl --> KCl + HOH 75.0 g 50.0 g ?g 75.0 g KOH mole KOH mole HCl 36.5 g HCl 56.1 g KOH mole KOH mole HCl = 48.8g HCl used 50.0 g - 48.8 = 1.2 g HCl left

Sample Problems 1) How many grams of carbon dioxide can be obtained from the action of 100.0 grams of hydrobromic acid on 100.0 grams of calcium carbonate? 2 HBr + CaCO3 --> HOH + CO2 + CaBr2 100.0 g 100.0 g ? g 100.0 g HBr mole HBr mole CO2 44.0 g CO2 80.9 g HBr 2 mole HBr mole CO2 = 27.2 g CO2 100.0 g CaCO3 mole CaCO3 mole CO2 44.0 g CO2 100.1 g CaCO3 mole CaCO3 mole CO2 = 44.0 g CO2

Sample Problems 2) How many grams of ammonia are evolved when 34 grams of ammonium chloride are added to 140 grams of barium hydroxide? 2 NH4Cl + Ba(OH)2 --> BaCl2 + 2 NH3 +2 HOH 34 g 140 g ? g 34 g NH4Cl mole NH4Cl 2 mole NH3 17.0 g NH3 53.5 g NH4Cl 2 mole NH4Cl mole NH3 = 11 g NH3 140 g Ba(OH)2 mole Ba(OH)2 2 mole NH3 17.0 g NH3 171.3 g Ba(OH)2 mole Ba(OH)2 mole NH3 = 28.0 gNH3

Sample Problems 3) 100.0 grams of lithium metal is dropped into 1.000 liter of water. How many grams of hydrogen are produced? 2 Li + 2 HOH --> 2 LiOH + H2 100.0 g 1000. g ? g 100.0 g Li mole Li mole H2 2.0 g H2 6.9 g Li 2 mole Li mole H2 = 14 g H2 1000. g HOH mole HOH mole H2 2.0 g H2 18.0 g HOH 2 mole HOH mole H2 = 56 g H2

Sample Problems 50.0 g O2 mole O2 mole C 12.0 g C 4) .50.0 grams of oxygen are available for the combustion of 25.0 grams of carbon. How many grams in excess is the oxygen or carbon ? O2 + C --> CO2 ? g excess 50.0 g 25.0 g 50.0 g O2 mole O2 mole C 12.0 g C 32.0 g O2 mole O2 mole C = 18.8g C used 25.0 g - 18.8 = 6.2 g C left

Sample Problems 5) 140.0 grams of sulfuric acid is added to 230.0 grams of barium peroxide. Which reactant is in excess and by how many grams? H2SO4 + BaO2 --> BaSO4 + H2O2 g excess 140. g 230. g 140.gH2SO4 mole H2SO4 mole BaO2 169.3 g BaO2 98.1 g H2SO4 mole H2SO4 mole BaO2 = 242 g BaO2 Since only 230. g available, BaO2 is the limiting reactant

Sample Problems = 133 g H2SO4 used 140. g - 133 = 7 g H2SO4 left 5) 140.0 grams of sulfuric acid is added to 230.0 grams of barium peroxide. Which reactant is in excess and by how many grams? H2SO4 + BaO2 --> BaSO4 + H2O2 g excess 140. g 230. g 230. g BaO2 mole BaO2 mole H2SO4 98.1 g H2SO4 169.3 g BaO2 mole BaO2 mole H2SO4 = 133 g H2SO4 used 140. g - 133 = 7 g H2SO4 left