Wednesday, March 27

Wednesday, March 27 EDMODO ASSIGNMENT

Watch the “Expanded Mole Map” video Draw the expanded map Click TURN IN and leave your answer to the

problem solved in the video as a comment. Print the PowerPoint notes for the unit. [You’ll

want the sample problem slides.] If needed, review the polyatomic ions, writing

equations and balancing equations.

Unit 7 StoichiometryChapter 9

FYI"Stoichiometry" is derived from the Greek words στοιχε ον (ῖ stoicheion, meaning element]) and με'τρον (metron, meaning measure.)Stoichiometry = measuring elements

Info in a Chemical Equation

2K + CuSO4 Cu + K2SO4

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 RelationshipsSince 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 O2

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.

2H2O 2H2 + O2

Example 9.2 (continued) What number of moles of O2 will

be produced by the decomposition of 5.8 mol of water?

2H2O 2H2 + O2 Next, the reaction shows that oxygen and

water have a quantitative relationship: a mole ratio.

1 mole O2 : 2 moles H2O 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 O2 will

be produced by the decomposition of 5.8 mol of water?

2H2O 2H2 + O2 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 9.3Goal: Solve the problem with a partner.

Calculate the number of moles of oxygen required to react exactly with 4.30 mol of propane, C3H8, in the reaction described by the following:

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

Calculate the number of moles of oxygen required to react exactly with

4.30 mol of propane, C3H8, in the reaction described by the following:

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

A.) 0.860 moles of O2

B.) 7.17 moles of O2

C.) 21.5 moles of O2

D.) none of these

Calculate the number of moles of oxygen required to react exactly with

4.30 mol of propane, C3H8, in the reaction described by the following:

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

A.) 0.860 moles of O2

B.) 7.17 moles of O2

C.) 21.5 moles of O2

D.) none of these

Example 9.4Is 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:

N2(g) + 3H2(g) = 2NH3(g)Calculate the number of moles of NH3 that can be made from 1.30 mol H2(g) reacting with excess N2(g).

Suggested Homework Page 287 4 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 9.5Goal: 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) + 3I2(s) = 2AlI3(s)Calculate the mass of I2(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) + 3I2(s) = 2AlI3(s)Calculate the mass of I2(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) + 3I2(s) = 2AlI3(s)Calculate the mass of I2(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.

493.89 g I2

Section 9.2 Review QuestionsGoal: 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) + CO2(g) Na2CO3(aq) + H2O(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) + CO2(g) Na2CO3(aq) + H2O(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 CO2

B.) 5.50 g CO2

C.) 9.09 g CO2

D.) none of these

NaOH(aq) + CO2(g) Na2CO3(aq) + H2O(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 CO2

B.) 5.50 g CO2

C.) 9.09 g CO2

D.) none of these

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 AlI3(s) formed by the reaction of 35.0 g Al(s) with 495 g I2.

Example 9.7Suppose 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 H2 is used? Left? How much N2 is used? Left?

Limiting Reactant Practice1.The reaction between solid white phosphorus and

oxygen produces solid tetraphosphorus decaoxide (P4O10). This compound is often called phosphorus pentoxide because its empirical formula is P2O5.

P4 + 5O2 P4O10

a. Determine the mass of tetraphosphorus decaoxide formed if 25.0 g of phosphorus (P4) 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?

Section 9.2 Review Questions4. You react 10.0 g of nitrogen gas

with hydrogen gas according to the following reaction.

N2(g) + 3H2(g) = 2NH3(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 N3- 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

The Plan: Stoichiometry Quiz (30 minute time

limit) Stoichiometry Stumpers

The lab practical will be similar to these examples.

5 point assignment due at the end of the period

Limiting Reactant Problem-based Learning 5 point assignment due at the end of the

period

Stoichiometry Test Tips1. 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) 100

7. 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 Cl2 gas if the products are NaCl and I2?

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 O2 if 7.36g of K2O is produced.