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Example of a mole to gram (2-step) stoichiometry problem

Example of a mole to gram (2-step) stoichiometry problem The Reaction: 6 CO 2 + 6 H 2 O  1 C 6 H 12 O 6 + 6 O 2 Question: If you have 5 moles of H 2 O how many grams of oxygen can you produce? Conversion: 5 moles H 2 O to grams of O 2.

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Example of a mole to gram (2-step) stoichiometry problem

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  1. Example of a mole to gram (2-step) stoichiometry problem The Reaction:6 CO2 + 6 H2O  1 C6H12O6 + 6 O2 • Question:If you have 5 moles of H2O how many grams of oxygen can you produce? • Conversion: 5 moles H2O to grams of O2 Start with the amount given in the question. Convert from moles to grams. Multiply by molar mass of O2 (16x2 = 32 g O2/mol) 6 molO2 6 molH2O = 5 molH2O X 5 molO2 X 160 g O2 32 g O2 1 mol O2 = This is the answer. Multiply by the mole ratio from the balanced equation with the unknown in the numerator. This cancels the given moles.

  2. Notes: What is percent yield? • Percent yield describes the efficiency of a chemical reaction. • It is a comparison between the actual yield and the theoretical yield. • Calculate the percent yield: • actual yield X 100% = percent yield • theoretical yield

  3. Calculate the percent yield: actual yield X 100% = percent yield theoretical yield You try 1: The theoretical yield is 200 g. The actual yield is 150 g. What is the percent yield? You try 2: The theoretical yield is 50 g. The actual yield is 25 g. What is the percent yield? You try 3: When 34 g ammonia are combined with excess oxygen, the theoretical yield of nitrogen gas is 28 g. The actual yield is 22 g. What is the percent yield of nitrogen gas? • Calculate the percent yield: actual yield X 100% = percent yield theoretical yield Example: The theoretical yield of silver when you combine 8g copper and 1.7g AgNO3 is 1.08 g Ag. The actual yield is 0.92 g Ag. What is the percent yield? 0.92g Ag X 100% = 85% 1.08 g Ag

  4. Notes: Percent Yield (to practice – p. 319 #6, p. 331 #34) • Example: hydrogen and oxygen form water 2 H2 (g) + O2 (g) → 2 H2O (l) • Identify the limiting reactant and the theoretical yield of water (grams) when 10 g hydrogen react with 32 g oxygen. The actual yield is 27g water. • What is the percent yield? Step 1: Split the question into two stoichiometry problems. Step 2: Solve both problems. Step 3: Compare answers. The smaller product amount is the theoretical yield. The limiting reactant is the reactant that that formed less product. Step 4: Percent yield – divide actual yield/theoretical x 100% • How many g H2O form from 10 g H2? • How many g H2O form from 32 g O2?

  5. Notes: Percent Yield (to practice – p. 319 #6, p. 331 #34) • Example: reaction of ammonia and oxygen 4 NH3 + 3 O2 → 6 H2O + 2 N2 • Identify the limiting reactant and the theoretical yield of Nitrogen (grams of N2) when 34 g ammonia (NH3 ) react with 64 g oxygen (O2). • The actual yield is 24 g N2. What is the percent yield? Step 1: Split the question into two stoichiometry problems. Step 2: Solve both problems. Step 3: Compare answers. The smaller product amount is the theoretical yield. The limiting reactant is the reactant that that formed less product. Step 4: Percent yield – divide actual yield/theoretical x 100% • How many grams of N2 form from 34 g NH3 ? • How many grams of N2 form from 64 g O2?

  6. N2 + 3 H2 2 NH3 • What is the mole ratio of H2 to NH3? • How many moles of H2 are needed to produce 10 moles NH3? • How many moles H2 are needed to produce 170 grams NH3? (molar mass of NH3 is 17g/mol) • How many grams H2 are needed to produce 170 grams NH3? (NH3 =17g/mol; H2 = 2g/mol) 3 molH2 2 mol NH3 3 molH2 2 mol NH3 = 10 mol NH3 X 15 mol H2 170g NH3 10 mol NH3 X 3 molH2 2 mol NH3 ÷ = 17 g/mol = 15 mol H2 3 molH2 2 mol NH3 170g NH3 ÷ = 15 mol H2 X 2 g/mol H2 17 g/mol = 10 mol NH3 X = 30 g H2

  7. Notes: Limiting Reactant & Product Yield • Limiting reactant = the substance that controls the quantity of product that can form in a chemical reaction. • The limiting reactant runs out first. • Excess reactant = the substance that is not used up completely in a reaction. • Product Yield = the amount of product formed • Theoretical Yield = amount of product predicted from calculations. • Think, Pair, Share: • What was the limiting reactant in your group’s s’more experiment? • What were the excess reactants?

  8. Notes: Limiting Reactant & Product Yield • Example 3: Copper and Silver Nitrate Cu (s) + 2AgNO3 (aq) → 2Ag (s) + Cu(NO3)2 (aq) • Identify the limiting reactant and the theoretical (calculated) yield of silver (Ag) in grams, when 128 g Cu reacts with 170 g AgNO3 . • Step 1: Separate the question into two stoichiometry problems: • How many grams Ag will be produced from 128 g Cu? • How many grams Ag will be produced from 170 g AgNO3? • Step 2: Solve both problems. • Step 3: Compare your answers. The smaller answer has the limiting reactant and is the theoretical yield.

  9. Notes: Limiting Reactant & Product Yield • Example 3: Copper and Silver Nitrate Cu (s) + 2AgNO3 (aq) → 2Ag (s) + Cu(NO3)2 (aq) • Identify the limiting reactant and the theoretical (calculated) yield of silver (Ag) in grams, when 128 g Cu reacts with 170 g AgNO3 . • How many grams Ag will be produced from 128 g Cu? • How many grams Ag will be produced from 170 g AgNO3?

  10. Practice: Limiting Reactant & Product Yield • Show all your work on your own sheet of paper. • Pre-Lab Problems: Copper and Silver Nitrate Cu (s) + 2AgNO3 (aq) → 2Ag (s) + Cu(NO3)2 (aq) • Identify the limiting reactant and the theoretical (calculated) yield of silver (Ag) in grams, when 64 g Cu reacts with 170 g AgNO3 . • Identify the limiting reactant and the theoretical (calculated) yield of silver (Ag) in grams, when 6.4 g Cu reacts with 8.5 g AgNO3 . • Identify the limiting reactant and the theoretical (calculated) yield of silver (Ag) in grams, when 32 g Cu reacts with 340 g AgNO3 . • Step 1: Separate the question into two stoichiometry problems: • Step 2: Solve both problems. • Step 3: Compare your answers. The smaller answer has the limiting reactant and is the theoretical yield.

  11. Check Your Understanding: 3e • Copper and Silver Nitrate Single Displacement Reaction: Cu (s) + 2AgNO3 (aq) → 2Ag (s) + Cu(NO3)2 (aq) • What is the mole ratio of silver (Ag) to copper (Cu) in the balanced equation? • What is the mole ratio of silver (Ag) to silver nitrate (AgNO3) in the balanced equation? • How many grams Ag will be produced from 640 g Cu? • How many grams Ag will be produced from 1700 g AgNO3? • Identify the limiting reactant and the theoretical (calculated) yield of silver (Ag) in grams, when 640 g Cu reacts with 1700 g AgNO3 .

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