CHAPTER 11 Stoichiometry 11.4 Solving Stoichiometric Problems
Section 11.1 Analyzing a Chemical Reaction A chemical equation tells us: What compounds are involved How much of each is used - Mole ratios can be determined using coefficients in a balanced equation C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(g) 1 mole glucose 2 moles ethanol 2 moles carbon dioxide
Section 11.1 Section 11.2 Analyzing a Chemical Reaction Percent Yield and Concentration - The percent yield tells us how much product has actually obtained, compared to the theoretical value. Obtained from the experiment Calculate using molar masses and mole ratios
Section 11.1 Section 11.2 Section 11.3 Analyzing a Chemical Reaction Percent Yield and Concentration Section 11.3 Limiting Reactants - When one reactant is completely used up, the whole reaction stops. - The reactant that is completely used up first is the limiting reactant. - If there is some reactant left over when the reaction stops, that reactant is the excess reactant. Excess reactant Limiting reactant
Section 11.1 Section 11.2 Section 11.3 Section 11.4 Analyzing a Chemical Reaction Section 11.2 Percent Yield and Concentration Section 11.3 Limiting Reactants Section 11.4 Solving Stoichiometric Problems Use what we’ve learned to answer these questions: - What is the limiting reactant? - What is the theoretical yield? What is the percent yield? - How much excess reactant is left? - How much reactant is used if it’s in a solution?
Section 11.1 Section 11.2 Section 11.3 Section 11.4 Analyzing a Chemical Reaction Section 11.2 Percent Yield and Concentration Section 11.3 Limiting Reactants Section 11.4 Solving Stoichiometric Problems Use what we’ve learned to answer these questions: - What is the limiting reactant? - What is the theoretical yield? What is the percent yield? - How much excess reactant is left? - How much reactant is used if it’s in a solution?
What is the limiting reactant? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: When 48.0 g of Li reacts with 46.5 g of N2, which reactant is the limiting reactant? Lithium is the only group 1 metal that is capable of reacting directly with nitrogen gas. 6Li(s) + N2(g) → 2Li3N(s)
What is the limiting reactant? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: When 48.0 g of Li reacts with 46.5 g of N2, which reactant is the limiting reactant? Lithium is the only group 1 metal that is capable of reacting directly with nitrogen gas. 6Li(s) + N2(g) → 2Li3N(s) Asked: Limiting reactant Given: 48.0 g of Li (have) 46.5 g of N2 (have)
What is the limiting reactant? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: When 48.0 g of Li reacts with 46.5 g of N2, which reactant is the limiting reactant? Lithium is the only group 1 metal that is capable of reacting directly with nitrogen gas. 6Li(s) + N2(g) → 2Li3N(s) Asked: Limiting reactant Given: 48.0 g of Li (have) 46.5 g of N2 (have) Relationships: molar mass of Li = 6.941 g/mole molar mass of N2 = 28.01 g/mole mole ratio: 6 moles Li ~ 1 mole N2
What is the limiting reactant? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: When 48.0 g of Li reacts with 46.5 g of N2, which reactant is the limiting reactant? Lithium is the only group 1 metal that is capable of reacting directly with nitrogen gas. 6Li(s) + N2(g) → 2Li3N(s) Asked: Limiting reactant Given: 48.0 g of Li (have) 46.5 g of N2 (have) Relationships: molar mass of Li = 6.941 g/mole molar mass of N2 = 28.01 g/mole mole ratio: 6 moles Li ~ 1 mole N2 Solve: 1) Moles of each reactant? 2) Apply the mole ratio 3) Compare what we have with what we need
What is the limiting reactant? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Limiting reactant Given: 48.0 g of Li (have) 46.5 g of N2 (have) Relationships: 6.941 g/mole Li 28.01 g/mole N2 6 moles Li ~ 1 mole N2 Solve: 1) Moles of each reactant? 2) Apply the mole ratio 3) Compare what we have with what we need
What is the limiting reactant? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Limiting reactant Given: 48.0 g of Li (have) 46.5 g of N2 (have) Relationships: 6.941 g/mole Li 28.01 g/mole N2 6 moles Li ~ 1 mole N2 Solve: 1) Moles of each reactant? 2) Apply the mole ratio 3) Compare what we have with what we need Have: 6.92 moles Li; 1.66 moles N2 How much N2 do we need to react with 6.92 moles Li?
What is the limiting reactant? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Limiting reactant Given: 48.0 g of Li (have) 46.5 g of N2 (have) Relationships: 6.941 g/mole Li 28.01 g/mole N2 6 moles Li ~ 1 mole N2 Solve: 1) Moles of each reactant? 2) Apply the mole ratio 3) Compare what we have with what we need Have: 6.92 moles Li; 1.66 moles N2 How much N2 do we need to react with 6.92 moles Li?
What is the limiting reactant? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Limiting reactant Given: 48.0 g of Li (have) 46.5 g of N2 (have) Relationships: 6.941 g/mole Li 28.01 g/mole N2 6 moles Li ~ 1 mole N2 Solve: 1) Moles of each reactant? 2) Apply the mole ratio 3) Compare what we have with what we need Have: 6.92 moles Li; 1.66 moles N2 Need: 1.15 moles N2 We need 1.15 moles N2 to react with 6.92 moles Li. Do we have enough N2?
What is the limiting reactant? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Limiting reactant Given: 48.0 g of Li (have) 46.5 g of N2 (have) Relationships: 6.941 g/mole Li 28.01 g/mole N2 6 moles Li ~ 1 mole N2 Solve: 1) Moles of each reactant? 2) Apply the mole ratio 3) Compare what we have with what we need Have: 6.92 moles Li; 1.66 moles N2 Need: 1.15 moles N2 We need 1.15 moles N2 to react with 6.92 moles Li. Do we have enough N2? Yes, we have more than enough N2. That means we will run out of Li before we run out of N2
What is the limiting reactant? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Limiting reactant Given: 48.0 g of Li (have) 46.5 g of N2 (have) Relationships: 6.941 g/mole Li 28.01 g/mole N2 6 moles Li ~ 1 mole N2 Solve: 1) Moles of each reactant? 2) Apply the mole ratio 3) Compare what we have with what we need Answer: Li is the limiting reactant Yes, we have more than enough N2. That means we will run out of Li before we run out of N2
Section 11.1 Section 11.2 Section 11.3 Section 11.4 Analyzing a Chemical Reaction Section 11.2 Percent Yield and Concentration Section 11.3 Limiting Reactants Section 11.4 Solving Stoichiometric Problems Use what we’ve learned to answer these questions: - What is the limiting reactant? - What is the theoretical yield? What is the percent yield? - How much excess reactant is left? - How much reactant is used if it’s in a solution?
What is the theoretical yield? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: How much lithium nitride (LiN3) can be produced from this reaction? 6Li(s) + N2(g) → 2Li3N(s)
What is the theoretical yield? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: How much lithium nitride (Li3N) can be produced from this reaction? 6Li(s) + N2(g) → 2Li3N(s) Asked: Amount of Li3N produced Given: Li is the limiting reactant 6.92 moles Li (have) Relationships: molar mass of Li3N = 34.83 g/mole mole ratio: 6 moles Li ~ 2 moles Li3N From the last problem
What is the theoretical yield? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: How much lithium nitride (Li3N) can be produced from this reaction? 6Li(s) + N2(g) → 2Li3N(s) Asked: Amount of Li3N produced Given: Li is the limiting reactant 6.92 moles Li (have) Relationships: molar mass of Li3N = 34.83 g/mole mole ratio: 6 moles Li ~ 2 moles Li3N Solve: 1) Find moles of Li3N 2) Convert moles to grams
What is the theoretical yield? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Amount of Li3N produced Given: Li is the limiting reactant 6.92 moles Li (have) Relationships: molar mass of Li3N = 34.83 g/mole mole ratio: 6 moles Li ~ 2 moles Li3N Solve: 1) Find moles of Li3N 2) Convert moles to grams
What is the theoretical yield? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Amount of Li3N produced Given: Li is the limiting reactant 6.92 moles Li (have) Relationships: molar mass of Li3N = 34.83 g/mole mole ratio: 6 moles Li ~ 2 moles Li3N Solve: 1) Find moles of Li3N 2) Convert moles to grams
What is the theoretical yield? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Amount of Li3N produced Given: Li is the limiting reactant 6.92 moles Li (have) Relationships: molar mass of Li3N = 34.83 g/mole mole ratio: 6 moles Li ~ 2 moles Li3N Solve: 1) Find moles of Li3N 2) Convert moles to grams Asked: 80.46 g of Li3N are produced
Section 11.1 Section 11.2 Section 11.3 Section 11.4 Analyzing a Chemical Reaction Section 11.2 Percent Yield and Concentration Section 11.3 Limiting Reactants Section 11.4 Solving Stoichiometric Problems Use what we’ve learned to answer these questions: - What is the limiting reactant? - What is the theoretical yield? What is the percent yield? - How much excess reactant is left? - How much reactant is used if it’s in a solution?
What is the percent yield? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: Calculate the percent yield of an experiment that actually produced 62.5 g of Li3N. 6Li(s) + N2(g) → 2Li3N(s)
What is the percent yield? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: Calculate the percent yield of an experiment that actually produced 62.5 g of Li3N. 6Li(s) + N2(g) → 2Li3N(s) Asked: Percent yield Given: Theoretical yield: 80.46 g Li3N Actual yield: 62.5 g Li3N Relationships: Solve: Use the percent yield formula From the last problem
What is the percent yield? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Percent yield Given: Theoretical yield: 80.46 g Li3N Actual yield: 62.5 g Li3N Relationships: Solve: Use the percent yield formula
What is the percent yield? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Percent yield Given: Theoretical yield: 80.46 g Li3N Actual yield: 62.5 g Li3N Relationships: Solve: Use the percent yield formula Answer: The percent yield in this particular experiment is 77.7%
Section 11.1 Section 11.2 Section 11.3 Section 11.4 Analyzing a Chemical Reaction Section 11.2 Percent Yield and Concentration Section 11.3 Limiting Reactants Section 11.4 Solving Stoichiometric Problems Use what we’ve learned to answer these questions: - What is the limiting reactant? - What is the theoretical yield? What is the percent yield? - How much excess reactant is left? - How much reactant is used if it’s in a solution?
What is left? 6Li(s) + N2(g) → 2Li3N(s) Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: How much of the excess reactant remains after the limiting reactant is completely consumed? 6Li(s) + N2(g) → 2Li3N(s)
What is left? 6Li(s) + N2(g) → 2Li3N(s) Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: How much of the excess reactant remains after the limiting reactant is completely consumed? 6Li(s) + N2(g) → 2Li3N(s) Asked: Amount of excess reactant left Given: N2 is the excess reactant 1.15 moles N2 (need) 1.66 moles N2 (have) Relationships: Molar mass of N2 = 28.01 g/mole From the first problem
What is left? 6Li(s) + N2(g) → 2Li3N(s) Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: How much of the excess reactant remains after the limiting reactant is completely consumed? 6Li(s) + N2(g) → 2Li3N(s) Asked: Amount of excess reactant left Given: N2 is the excess reactant 1.15 moles N2 (need) 1.66 moles N2 (have) Relationships: Molar mass of N2 = 28.01 g/mole Solve: 1) How many moles N2 remain? 2) Convert moles to grams
What is left? 6Li(s) + N2(g) → 2Li3N(s) Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Amount of excess reactant left Given: N2 is the excess reactant 1.15 moles N2 (need) 1.66 moles N2 (have) Relationships: 28.01 g/mole N2 Solve: 1) How many moles N2 remain? 2) Convert moles to grams
What is left? 6Li(s) + N2(g) → 2Li3N(s) Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Amount of excess reactant left Given: N2 is the excess reactant 1.15 moles N2 (need) 1.66 moles N2 (have) Relationships: 28.01 g/mole N2 Solve: 1) How many moles N2 remain? 2) Convert moles to grams
What is left? 6Li(s) + N2(g) → 2Li3N(s) Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Amount of excess reactant left Given: N2 is the excess reactant 1.15 moles N2 (need) 1.66 moles N2 (have) Relationships: 28.01 g/mole N2 Solve: 1) How many moles N2 remain? 2) Convert moles to grams Answer: 14 g of N2 will remain at the end of the reaction.
Section 11.1 Section 11.2 Section 11.3 Section 11.4 Analyzing a Chemical Reaction Section 11.2 Percent Yield and Concentration Section 11.3 Limiting Reactants Section 11.4 Solving Stoichiometric Problems Use what we’ve learned to answer these questions: - What is the limiting reactant? - What is the theoretical yield? What is the percent yield? - How much excess reactant is left? - How much reactant is used if it’s in a solution?
Reactants in solution CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. What is the concentration of CuSO4 in the water if 0.021 g of CuS precipitate is formed? The reaction is: CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
Reactants in solution CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. What is the concentration of CuSO4 in the water if 0.021 g of CuS precipitate is formed? The reaction is: CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) Asked: Concentration of CuSO4(aq) Given: 1.0 L of solution is tested 0.021 g CuS (formed) Relationships: Molar mass of CuS = 95.61 g/mole Mole ratio: 1 mole CuSO4 ~ 1 mole CuS
Reactants in solution CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. What is the concentration of CuSO4 in the water if 0.021 g of CuS precipitate is formed? The reaction is: CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) Asked: Concentration of CuSO4(aq) Given: 1.0 L of solution is tested 0.021 g CuS (formed) Relationships: Molar mass of CuS = 95.61 g/mole Mole ratio: 1 mole CuSO4 ~ 1 mole CuS Solve: 1) How many moles of CuS? 2) How many moles of CuSO4? 3) What is the concentration of CuSO4?
Reactants in solution CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is: CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) Asked: Concentration of CuSO4(aq) Given: 1.0 L of solution is tested 0.021 g CuS (formed) Relationships: 95.61 g/mole CuS 1 mole CuSO4 ~ 1 mole CuS Solve: 1) How many moles of CuS? 2) How many moles of CuSO4? 3) What is the concentration of CuSO4?
Reactants in solution CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is: CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) Asked: Concentration of CuSO4(aq) Given: 1.0 L of solution is tested 0.021 g CuS (formed) Relationships: 95.61 g/mole CuS 1 mole CuSO4 ~ 1 mole CuS Solve: 1) How many moles of CuS? 2) How many moles of CuSO4? 3) What is the concentration of CuSO4? Have:
Reactants in solution CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is: CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) Asked: Concentration of CuSO4(aq) Given: 1.0 L of solution is tested 0.021 g CuS (formed) Relationships: 95.61 g/mole CuS 1 mole CuSO4 ~ 1 mole CuS Solve: 1) How many moles of CuS? 2) How many moles of CuSO4? 3) What is the concentration of CuSO4? Have:
Reactants in solution CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is: CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) Asked: Concentration of CuSO4(aq) Given: 1.0 L of solution is tested 0.021 g CuS (formed) Relationships: 95.61 g/mole CuS 1 mole CuSO4 ~ 1 mole CuS Solve: 1) How many moles of CuS? 2) How many moles of CuSO4? 3) What is the concentration of CuSO4? Answer: The concentration of CuSO4 is 2.20 x 10-4 M.
Reactants in solution CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is: CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) Discussion: If the legal limit is 5.0 x 10-4 M of Cu2+, is the industrial plant following the environmental guidelines? Answer: The concentration of CuSO4 is 2.20 x 10-4 M.
Reactants in solution CuSO4(aq) → Cu2+(aq) + SO42–(aq) The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is: CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) Discussion: If the legal limit is 5.0 x 10-4 M of Cu2+, is the industrial plant following the environmental guidelines? Yes, because 2.20 x 10-4 M is less than the legal limit. Answer: The concentration of CuSO4 is 2.20 x 10-4 M. CuSO4(aq) → Cu2+(aq) + SO42–(aq)
Section 11.1 Section 11.2 Section 11.3 Section 11.4 Analyzing a Chemical Reaction Section 11.2 Percent Yield and Concentration Section 11.3 Limiting Reactants Section 11.4 Solving Stoichiometric Problems Use what we’ve learned to answer these questions: - What is the limiting reactant? - What is the theoretical yield? What is the percent yield? - How much excess reactant is left? - How much reactant is used if it’s in a solution?