As you come in,  The Materials:  Remote control  Pick up packet.  Paper, pencil, calculator, periodic table for notes  The Plan:  Learn about stoichiometry  The Assessments:  Stoichiometry Quiz on Thursday  Limiting Reactants Quiz on Friday Halloween Pandora Station – Project Eliminate Challenge

Unit 7 Stoichiometry Chapter 9

Info in a Chemical Equation 2K + CuSO 4  Cu + K 2 SO 4  A balanced chemical equation shows the correct ratios required for a chemical reaction to occur.  Knowing the ratio allows us to provide the appropriate amount of reactants AND predict the amount of products.

Mole to Mole Relationships  Since chemical equations give appropriate relationships of moles of each compound, mole ratios can be written.  Mole ratio : uses the coefficients in the balanced equation to show how two compounds are related in a reaction

Example 9.2 Goal : Write a mole ratio & solve the problem.  What number of moles of O 2 will be produced by the decomposition of 5.8 mol of water?  First, the balanced chemical equation must be written for the situation described in the prompt.  2H 2 O  2H 2 + O 2

Example 9.2 (continued)  What number of moles of O 2 will be produced by the decomposition of 5.8 mol of water? 2H 2 O  2H 2 + O 2  Next, the reaction shows that oxygen and water have a quantitative relationship: a mole ratio. 1 mole O 2 : 2 moles H 2 O  Where did the numbers (1 & 2) come from?  Can you write other mole ratios from this equation?

Example 9.2 (continued)  What number of moles of O 2 will be produced by the decomposition of 5.8 mol of water? 2H 2 O  2H 2 + O 2  Finally, we can solve this problem using the mole ratio as a conversion factor in our dimensional analysis.  Start with the given.  Set up the chart to cancel units AND compounds.  Multiply or divide as usual.

Example 9.3 Goal : Solve the problem with a partner.  Calculate the number of moles of oxygen required to react exactly with 4.30 mol of propane, C 3 H 8, in the reaction described by the following: C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g)

Calculate the number of moles of oxygen required to react exactly with 4.30 mol of propane, C 3 H 8, in the reaction described by the following: C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g) A.) moles of O 2 B.) 7.17 moles of O 2 C.) 21.5 moles of O 2 D.) none of these

Calculate the number of moles of oxygen required to react exactly with 4.30 mol of propane, C 3 H 8, in the reaction described by the following: C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g) A.) moles of O 2 B.) 7.17 moles of O 2 C.) 21.5 moles of O 2 D.) none of these

Example 9.4 Is another example needed?  Ammonia is used in huge quantities as a fertilizer. It is manufactured by combining nitrogen and hydrogen according to the following equation: N 2 (g) + 3H 2 (g) = 2NH 3 (g) Calculate the number of moles of NH 3 that can be made from 1.30 mol H 2 (g) reacting with excess N 2 (g).

Suggested Homework  Page and 5

Mega Mole Map  Mole to mole relationship  Mole Ratio = conversion factor (bridge)  In our examples so far, we’ve been given moles of one compound and asked to convert to moles of another compound.  What if I told you that I won’t always start out with moles or ask for moles? Starting or ending with mass, liters, or particles is very common.  Work with your partner to expand on our current Mole Concept Map.

Mega Mole Map  Mole to mole relationship  Mole Ratio = conversion factor (bridge)  Mole to mass relationship  Molar Mass = conversion factor (bridge)  Mole to particle relationship  Avogadro’s # = conversion factor (bridge)  Mole to volume relationship  Molar volume = conversion factor (bridge)

Example 9.5 Goal : Use new mole concepts to solve the problem.  Consider the reaction of powdered aluminum metal and finely ground iodine to produce aluminum iodide. The balanced equation for this vigorous chemical reaction is: 2Al(s) + 3I 2 (s) = 2AlI 3 (s) Calculate the mass of I 2 (s) needed to just react with 35.0 g of Al(s).  First, write the given and set up your chart.

Example 9.5 (continued)  Consider the reaction of powdered aluminum metal and finely ground iodine to produce aluminum iodide. The balanced equation for this vigorous chemical reaction is: 2Al(s) + 3I 2 (s) = 2AlI 3 (s) Calculate the mass of I 2 (s) needed to just react with 35.0 g of Al(s).  Next, consult your map to determine a problem- solving path. How many steps will this problem require?  3 steps

Example 9.5 (continued)  Consider the reaction of powdered aluminum metal and finely ground iodine to produce aluminum iodide. The balanced equation for this vigorous chemical reaction is: 2Al(s) + 3I 2 (s) = 2AlI 3 (s) Calculate the mass of I 2 (s) needed to just react with 35.0 g of Al(s).  Finally, plug in your conversion factors. Multiply and divide as usual to give your final answer.  g I 2

Section 9.2 Review Questions Goal : Solve the problem with a partner. 2. Solutions of sodium hydroxide cannot be kept for very long because they absorb carbon dioxide from the air, forming sodium carbonate. The unbalanced equation is: NaOH(aq) + CO 2 (g)  Na 2 CO 3 (aq) + H 2 O(l) Calculate the number of grams of carbon dioxide that can be absorbed by complete reaction with a solution that contains 5.00 g of NaOH.

NaOH(aq) + CO 2 (g)  Na 2 CO 3 (aq) + H 2 O(l) Calculate the number of grams of carbon dioxide that can be absorbed by complete reaction with a solution that contains 5.00 g of NaOH. A.) 2.75 g CO 2 B.) 5.50 g CO 2 C.) 9.09 g CO 2 D.) none of these

NaOH(aq) + CO 2 (g)  Na 2 CO 3 (aq) + H 2 O(l) Calculate the number of grams of carbon dioxide that can be absorbed by complete reaction with a solution that contains 5.00 g of NaOH. A.) 2.75 g CO 2 B.) 5.50 g CO 2 C.) 9.09 g CO 2 D.) none of these

Suggested Homework  11-2 Practice Problems – (1-11…skip 9) NOTE : We’ll use iRespond TODAY to check your answers, and we’ll make colored pencil corrections on your work.

Limiting Reactants  As you know, a balanced chemical equation gives the perfect ratio of reactants needed to perform the reaction.  In reality, we rarely have the “perfect” ratio.  With an imperfect ratio, one reactant will run out before the other. The reactant running out will stop the reaction...limit the products.  The reactant that runs out = Limiting reactant  The reactant that remains = Excess reactant

Limiting Reactant Calculations  Using stoichiometry, we can calculate the following predictions:  Identify the limiting reactant  Identify the excess reactant  Predict the amount of product to form  Predict the amount of excess reactant remaining

Practice Problem 9.5  Calculate the mass of AlI 3 (s) formed by the reaction of 35.0 g Al(s) with 495 g I 2.

Example 9.7  Suppose that 25.0 kg of nitrogen gas and 5.00 kg of hydrogen gas are mixed and reacted to form ammonia. Calculate the mass of ammonia produced when this reaction is run to completion.  Two “givens” means two calculations  The unknown is the same in the two calculations  The smallest answer is the best prediction as it tells when the limiting reactant will run out.

Follow-up Questions  Identify the limiting reactant.  Identify the excess reactant.  How much H 2 is used? Left?  How much N 2 is used? Left?

Limiting Reactant Practice 1.The reaction between solid white phosphorus and oxygen produces solid tetraphosphorus decaoxide (P 4 O 10 ). This compound is often called phosphorus pentoxide because its empirical formula is P 2 O 5. P 4 + 5O 2  P 4 O 10 a. Determine the mass of tetraphosphorus decaoxide formed if 25.0 g of phosphorus (P 4 ) and 50.0 g of oxygen gas are combined. b. Identify the limiting reactant. c. How much of the excess reactant remains after the reaction stops?

FYI  "Stoichiometry" is derived from the Greek words στοιχε ῖ ον (stoicheion, meaning element]) and μέτρον (metron, meaning measure.)

Section 9.2 Review Questions 4. You react 10.0 g of nitrogen gas with hydrogen gas according to the following reaction. N 2 (g) + 3H 2 (g) = 2NH 3 (g) a. What mass of hydrogen is required to completely react with 10.0 g sample of nitrogen gas? b. What mass of ammonia is produced from 10.0 g of nitrogen gas and sufficient hydrogen gas?

Practice Problem 9.8  Lithium nitride, an ionic compound containing the Li + and N 3- ions, is prepared by the reaction of lithium metal and nitrogen gas. Calculate the mass of lithium nitride formed from 56.0 g of nitrogen gas and 56.0 g of lithium.

Follow-up Questions  Identify the limiting reactant.  Identify the excess reactant.  How much of each reactant is used? Left?

Percent Yield  Yield means product.  Calculating % yield is calculating what % of the product your experiment actually produced.  You are comparing your prediction to your action.

% Yield  (Actual yield / theoretical yield) 100  Actual yield = either given in the problem or carried out in the lab  Theoretical yield = ALWAYS calculated by stoichiometry

Stoichiometry Test Tips 1. Know how to write and balance a combustion reaction. (See #1 on your Stoichiometry Quiz.) 2. Remember, “a liter” or “a single gram” means 1 L or 1 gram. The given can be written this way. 3. “How many grams of product are produced?” - refers to theoretical yield (smallest answer) in a limiting reactant problem

4. Metal products are solids. 5. STP = Standard temperature and pressure (doesn’t affect the calculation) 6. % yield = (ACTUAL/THEORETICAL) Excess reactant - Use dimensional analysis starting with given amount of LR to calculate the amount USED; Subtract from ORIGINAL amount to calculate excess LEFT

Look closely at the wording…  Labeling the equation will help!  If you are given info about BOTH reactants, then you have a limiting reactant problem.  Example: What mass of NaCl will be produced by the reaction of 58.7g of NaI with 29.4g of Cl 2 gas if the products are NaCl and I 2 ?  If you are given info about ONLY ONE substance, then you have a simple stoichiometry problem.

More wording advice…  LABELS!  If you are given info about BOTH reactants AND a product, then you’ve probably got a percent yield problem to solve. The product info is the actual yield.  Determine the percent yield for a reaction between 6.92g K and 4.28g of O 2 if 7.36g of K 2 O is produced.