Stoichiometry Interpreting Balanced Equations Stoichiometrical Calculations Limiting Reagents Percent Yield

Stoichiometry Stoichiometry is the study of the amount of substances produced and consumed in chemical reactions. The word stoichiometry comes from the Greek “stoicheion” and “metrein” meaning “element” and to “measure”. Calculations are carried out step-by-step using dimensional analysis.

Interpreting Chemical Equations A balanced equation shows the ratios in which the substances combine.

2Al(s) + 6HCl(aq) → 2AlCl3(s) + 3H2(g) atoms 6 molecule 2 Formula units 3 molecule 2 mole 6 moles 2 mole 3 moles 3(22.4) = 67.2 LSTP LSTP LSTP LSTP 2(26.98 g) = 53.96 6(36.46 g) =218.76 2(133.33 g) = 266.66 3(2.02 g) =6.06 g g g g What is true about the mass of the products and the mass of the reactants? They are equal. 53.96 g + 218.76 g = 266.66 g + 6.06 g

2Al(s) + 6HCl(aq) → 2AlCl3(s) + 3H2(g) atoms 6 molecule 2 Formula units 3 molecule 2 mole 6 moles 2 mole 3 moles 3(22.4) = 67.2 LSTP LSTP LSTP LSTP 2(26.98 g) = 53.96 6(36.46 g) =218.76 2(133.33 g) = 266.66 3(2.02 g) =6.06 g g g g According to the balanced equation, if two moles of aluminum react with 6 moles of hydrochloric acid, how many moles of aluminum chloride will be produced? How many moles of hydrogen gas will be produced? 2 mol AlCl3 3 mol H2

2Al(s) + 6HCl(aq) → 2AlCl3(s) + 3H2(g) atoms 6 molecule 2 Formula units 3 molecule 2 mole 6 moles 2 mole 3 moles 3(22.4) = 67.2 LSTP LSTP LSTP LSTP 2(26.98 g) = 53.96 6(36.46 g) =218.76 2(133.33 g) = 266.66 3(2.02 g) =6.06 g g g g How would the amount of products produced be different if twice as many moles of each of the reactants were used? The amount of products produced would also be doubled (4 moles of AlCl3, 6 moles of H2)

2Al(s) + 6HCl(aq) → 2AlCl3(s) + 3H2(g) A mole ratio is the ratio between the numbers of moles of any two substances in a balanced chemical equation. Write the possible mole ratios for the reaction. 2 mol Al : 6 mol HCl 2 mol Al : 2 mol AlCl3 2 mol Al : 3 mol H2 6 mol HCl : 2 mol AlCl3 6 mol HCl : 3 mol H2 2 mol AlCl3 : 3 mol H2

Mole ratios can be written as conversion factors. Examples: The mole ratio 2 mol Al: 6 mol HCl Can be written as 2 mol Al 6 mol HCl or 6 mol HCl 2 mol Al The mole ratio 6 mol HCl: 3 mol H2 Can be written as 6 mol HCl 3 mol H2 or 3 mol H2 6 mol HCl

Stoichiometrical Calculations A balanced chemical equation is necessary in order to do stoichiometrical calculations The Mole Bridge aG  bW (given quantity) (wanted quantity) Mass of G = Mass of W mol of G mol of W Volume of G (liters) at STP Volume of W (liters) at STP

Example Problems. Remember to balance your Equations! 1. How many moles of nitrogen gas are required to completely react with 4.5 moles of hydrogen gas to produce ammonia? N2 + 3H2 → 2NH3

2. Sodium chloride is prepared by the reaction of sodium metal with chlorine gas. How many moles of sodium are required to produce 40.0 g of sodium chloride? 2Na + Cl2 → 2NaCl

3. How many grams of aluminum will be produced by the decomposition of 25.0 g of aluminum oxide? 2Al2O3 → 4Al + 3O2

4. Oxygen is prepared in the laboratory by the decomposition of potassium chlorate. Calculate the volume of oxygen in liters, measured at STP, that would be obtained from 183.9 g of potassium chlorate. 2KClO3 → 2KCl + 3O2

5. Calculate the volume of carbon dioxide that will be produced at STP from the complete combustion of 6.0 L of ethane gas (C2H6). 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)

6. If 30.0 g of lithium react with an excess of water, how many grams of lithium hydroxide will be produced? 2Li + 2H2O → 2LiOH + H2

Limiting Reagent The limiting reagent in a chemical reaction limits the amounts of the other reactants that can combine – and the amount of product that can form – in a chemical reaction. The reaction will stop once the limiting reagent runs out.

Excess Reagent The excess reagent in a chemical reaction is the substance that is not used up completely in a chemical reaction.

An Example Mrs. Jones likes to make grilled cheese sandwiches for her children. 2 slices bread + 1 slice cheese + 1 T Butter → 1 sandwich How many sandwiches could Mrs. Jones make if she has 10 slices of bread, 4 slices of cheese and 7 tablespoons of butter? Why? Mrs. Jones could only make 4 sandwiches because she only has enough cheese for 4 sandwiches. What left over ingredients would Mrs. Kyle have? Mrs. Jones would have 2 slices of bread and 3 tablespoons of butter left over.

Ammonia is prepared by the reaction of hydrogen gas with nitrogen gas. 3H2 + N2 → 2NH3 Determine the maximum amount of ammonia (in moles) that could be produced when 6.0 mol of H2 reacts with 4.0 mol of N2 and identify the limiting reagent. 6.0 mol H 2 × 2 mol NH3 3 mol H2 =4.0 mol N H 3 4.0 mol N 2 × 2 mol NH3 1 mol N 2 =8.0 mol NH 3 The maximum amount of NH3 that could be produced is 4.0 mol and H2 is limiting.

Ammonia is prepared by the reaction of hydrogen gas with nitrogen gas. 3H2 + N2 → 2NH3 How many moles of the excess reagent remain unreacted? 6.0 mol H 2 × 1 mol N 2 3 mol H2 =2.0 mol N 2 reacted 4.0 mol N2 – 2.0 mol N2 = 2.00 mol N2 unreacted

Ammonia is prepared by the reaction of hydrogen gas with nitrogen gas. 3H2 + N2 → 2NH3 Draw a representation of the reaction before and after the reaction. H2 N2 NH3 Before Reaction After Reaction

Methanol, CH3OH, is the simplest of the alcohols Methanol, CH3OH, is the simplest of the alcohols. It is synthesized by the reaction of hydrogen and carbon monoxide. 2H2 + CO → CH3OH If 150 mol of carbon monoxide and 400 moles of hydrogen chemically combine, determine the maximum amount (in mol) of methanol that could be produced and identify the limiting reagent. The maximum amount of CH3OH that could be produced is 150 mol and CO is limiting.

2H2 + CO → CH3OH 400 mol H2 – 300 mol H2 = 100 mol H2 Methanol, CH3OH, is the simplest of the alcohols. It is synthesized by the reaction of hydrogen and carbon monoxide. 2H2 + CO → CH3OH How many mol of the excess reagent remain unreacted? 400 mol H2 – 300 mol H2 = 100 mol H2

A student reacts 80. 0 g of copper metal with 25 A student reacts 80.0 g of copper metal with 25.0 g of sulfur to form copper(i) sulfide. 2Cu + S → Cu2S Determine the maximum amount of copper(I) sulfide (in grams) that could be produced and identify the limiting reagent. The maximum amount of Cu2S that can be produced is 100. g. The limiting reagent is copper.

A student reacts 80. 0 g of copper metal with 25 A student reacts 80.0 g of copper metal with 25.0 g of sulfur to form copper(i) sulfide. 2Cu + S → Cu2S How many grams of the excess reagent remain unreacted? 25.00 g S – 20.19 g S = 4.81g S

Percent Yield The theoretical yield is the calculated amount of product that could form during a reaction based upon a balanced equation. This is the maximum amount of product that could be formed.

Percent Yield The actual yield is the amount of product that forms when the reaction is carried out in the laboratory. The actual yield is often less than the theoretical yield.

Percent Yield The percent yield is the ratio of the actual yield to the theoretical yield.

What are some factors that might cause the percent yield to be less than 100%? Reactions do not always go to completion. Impure reactants and competing side reactions may cause other products to be formed. Some of the product may be lost during purification.

What are some factors that might cause the percent yield to appear to be more than 100%? The product might not be completely dry. The product might combine with oxygen in the air drying it.

A student decomposed 24.8 g calcium carbonate in the laboratory. CaCO3 → CaO + CO2 Calculate the theoretical yield of calcium oxide. What is the percent yield if the student only produced 13.1 g of calcium oxide in the laboratory?

SiO2(s) + 3C(s) → SiC(s) + 2CO(g) When 50.0 g of silicon dioxide is heated with an excess of carbon, 32.2 g of silicon carbide is produced. SiO2(s) + 3C(s) → SiC(s) + 2CO(g) Calculate the theoretical yield. Calculate the percent yield.

2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s) Aluminum reacts with excess copper(II) sulfate according to the reaction given below. If 1.95 g of Al react and the percent yield of Cu is 60.0%, what mass of cu is produced? 2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s) First calculate the theoretical yield of copper. Next calculate the mass of Cu produced.

2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s) Aluminum reacts with excess copper(II) sulfate according to the reaction given below. If 1.95 g of Al react and the percent yield of Cu is 60.0%, what mass of Cu is produced? 2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s) First calculate the theoretical yield of copper. Next calculate the mass of Cu produced.