Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic...

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Stoichiometry

Transcript of Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic...

Page 1: Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic molecules using the yellow balls as well. Without.

Stoichiometry

Page 2: Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic molecules using the yellow balls as well. Without.

Inquiry Activity• Using 20 white balls (H), make 10 diatomic

molecules. Make diatomic molecules using the yellow balls as well.

• Without looking, have someone choose 10 molecules at random.

• Out of the chosen molecules, pair up the reactants using the following chemical equation: Y2 + 3W2 2YW3

• Pair up reactants until you run out of one of the reactants. This is your limiting reagent.

• Repeat the activity and see if you get the same results

Page 3: Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic molecules using the yellow balls as well. Without.

Review – Balance these equations

1. N2 + H2 NH3

2. H2S + O2 SO2 + H2O

3. H2 + I2 HI

Page 4: Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic molecules using the yellow balls as well. Without.

Stoichiometry• Study of the amount of substances consumed

or produced in a chemical reaction– Requires Balanced Equations!

Reactants Products

Page 5: Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic molecules using the yellow balls as well. Without.

StoichiometryReactants and Products

ALWAYS the same

• Number of Atoms• Mass

Reactants and Products SOMETIMES the same

• Number of molecules• Moles• Volume

N2 + 3H2 2NH3

Atoms 2 N 6 H = 2 N + 6 H Mass 28g + 6g = 34gNumber ofmolecules 1 molecule 3 molecules 2 molecules Moles 1mole 3 moles 2molesVolume 22.4L 64.2L 44.8L

Page 6: Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic molecules using the yellow balls as well. Without.

What is conserved in each reaction?Atoms? Mass? Number of Molecules? Moles?

1. H2 + O2 H2O

2. H2 + I2 2HI

3. C2H2 + O2 CO2 + H2O

Page 7: Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic molecules using the yellow balls as well. Without.

Calculations• Mole ratio: Conversion factor using the coefficients

of balanced chemical equations– Used to convert between moles of reactant and moles

of product

N2 + 3H2 2NH3

1 mol 3 mol 2mol

So…..1 mol N2 2 mol NH3 3mol H2

3 mol H2 1 mol N2 2 mol NH3

Page 8: Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic molecules using the yellow balls as well. Without.

• How many moles of ammonia (NH3) are produced when 0.6mol nitrogen (N2) reacts with hydrogen (H2)?1. Write the chemical reaction:

2. What conversion do you use to get from N2 to NH3?

3. Solve

Page 9: Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic molecules using the yellow balls as well. Without.

1. Al + O2 Al2O3

How many moles of aluminum are needed to form 3.7mol of aluminum oxide?

How many moles of oxygen are required to react completely with 14.8mol Al?

How many moles of Al2O3 are formed when 0.78 mol O2 reacts with aluminum?

Page 10: Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic molecules using the yellow balls as well. Without.

Mole – Mass Calculations• Calculate the number of grams of NH3 produced

by the reaction of 5.40g hydrogen with an excess of nitrogen.1. What is the chemical reaction?

2. Conversiongram hydrogen mol hydrogen mol NH3 gram NH3

Page 11: Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic molecules using the yellow balls as well. Without.

CaC2 + H2O C2H2 + Ca(OH)2

1. How many grams of acetylene gas (C2H2) are produced by adding water to 5.00g CaC2?

2. How many moles of CaC2 are needed to react completely with 49.0g H2O?

Page 12: Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic molecules using the yellow balls as well. Without.

Other conversions…• Number of particles - 1 mol

6.02x1023 particles

• Volume of gas at STP - 1 mol22.4L

So you should be able to convert:mass-mass mol-molmass-mol mol-number of particlesmol-volume of gas at STP

ALYWAYS CONVERT THROUGH USING MOLES!!!

Page 13: Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic molecules using the yellow balls as well. Without.

1. How many molecules of oxygen are produced when 29.2g of water is decomposed?

H2O H2 + O2

2. How many grams of nitrogen dioxide must react with water to produce 5.0x1022 molecules of nitrogen monoxide?

NO2 + H2O HNO3 + NO

Page 14: Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic molecules using the yellow balls as well. Without.

• How many liters of nitrogen dioxide are produced when 34L of oxygen reacts with excess nitrogen monoxide?

NO + O2 NO2

Page 15: Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic molecules using the yellow balls as well. Without.

How much do you actually get?• In any reaction, the amount of product formed is

limited by the amount of reagents….– Limiting Reagent: -determines the amount of product

that can be formed -is completely used up in the

reaction

– Excess Reagent: - reactant is not completely used up

Page 16: Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic molecules using the yellow balls as well. Without.

• What is the limiting reagent when 80g of Cu reacts with 25g of S in the following reaction?

Cu + S Cu2S1.Balance the equation2.Convert grams to moles

3.What is the mole ratio between Cu and S?

4.Take the given mole quantity for one reagent (#2), multiply it by the mole ratio (#3) to find how many moles of the other reagent are needed

5.Compare moles needed to moles you have to find which one is limiting

Page 17: Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic molecules using the yellow balls as well. Without.

1. If 2.70mol of C2H4 is reacted with 6.30mol O2, what is the limiting reagent?

C2H4 + O2 CO2 + H2O

2. What is the limiting reagent if 6.0g HCl reacts with 5.0g Mg?

Mg + HCl MgCl2 + H2

Page 18: Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic molecules using the yellow balls as well. Without.

Limiting reagents can be used to find out how much product is actually made.

• What is the maximum number of grams of Cu2S that can be formed when 80.0 g Cu reacts with 25.0g S?

Cu + S Cu2S

1.What is the limiting reagent?2.Use the limiting reagent to calculate the amount of product

Page 19: Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic molecules using the yellow balls as well. Without.

• How many grams of water can be produced by the reaction of 2.40mol C2H2 with 7.4mol O2?

C2H2 + O2 CO2 + H2O

• If 2.7mol C2H4 is reacted with 6.3mol O2, how many moles of water are produced?

C2H4 + O2 CO + H2O

Page 20: Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic molecules using the yellow balls as well. Without.

Things happen in lab…

• Theoretical yield: the maximum amount of product that could be formed

• Actual yield: the amount of product that is actually formed in lab

Percent Yield = Actual yield x 100% Theoretical yield

Page 21: Stoichiometry. Inquiry Activity Using 20 white balls (H), make 10 diatomic molecules. Make diatomic molecules using the yellow balls as well. Without.

• What is the percent yield if 13.1g CaO is actually produced when 24.8g CaCO3 is heated?

CaCO3 CaO + CO2