Chemical Equations and Reaction Stoichiometry

Chapter 3

Chemical Reactions

Reactant Reactant Product Product

1 molecule CH4

2 molecules O2

1 molecule CO2

2 molecules H2O

1 mole CH4 2 moles O2 1 mole CO2 2 moles H2O

16.0 grams CH4

64.0 grams O2 (2 32.0 g)

44.0 g CO2 36.0 grams H2O (2 18.0

g)

Physical State Representation

• (g) = _________

• (l) = _________

• (s) = _________

• (aq) = _______________

These representations will commonly be observed in chemical equations.

Chemical Equations

• There is no loss in the quantity of matter from a chemical reaction. This is the _____ __ __________ __ _____.

• What is the total mass on both sides of the chemical reaction shown previously?

• A balanced chemical equation must have the _____ of each kind of atom on both sides of the equation.– Smallest possible whole-number coefficients.

Balancing Chemical Equations

• __Al(s) + __Br2(l) ___Al2Br6(s) (covalent)

• __C3H8(g) + __O2(g) __CO2(g) + __H2O(g)

• __C4H10(l) + __O2(g) __CO2(g) + __H2O(g)

• __P4(s) + __Cl2(g) __PCl3(s)– Demo: On demo. CD

Be sure to balance equations before proceeding with calculations.

Calculations Based on Balanced Chemical Equations

• The ______ in a balanced equation represent the relative ratio of moles or molecules (formula units).

• Calculations utilize a series of _________ to obtain the desired answer.– Make sure that the ________ cancel.

Calculations Based on Balanced Chemical Equations

• Conversion process– Convert grams to moles (using molar mass)– Convert moles to moles (from balanced

equation)• This can also be molecules. Why?

– Convert moles to grams (using molar mass)

This is the general steps that will be followed to solve equations.

Solving Problems from Chemical Equations

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

• How many CH4 molecules required to react completely with 82 molecules of O2?

__C4H10(l) + __O2(g) __CO2(g) + __H2O(g)

• How many O2 molecules required to react completely with 48 molecules of C4H10?

Solving Problems from Chemical Equations

__Fe2O3(s) + __CO(g) __Fe(s) + __CO2(g)• How many moles of Fe2O3 are required to react completely

with 4.9 moles of CO?• What mass of CO2 can be produced from 0.540 moles of

Fe2O3 with excess CO?

__P4(s) + __Cl2(g) __PCl3(s)• How many moles of PCl3 can be produced from 1.48

moles of Cl2(g)? Assume the P4 is in excess.• How many grams of PCl3 can be produced from 42.3

grams of P4?

More Problems??

__NH3 + __O2 __NO + __H2O

• How many grams of NO can be produced from 17.80 grams of O2? NH3 is in excess.

• How many molecules of NH3 are required to produce 7.31 10-10 grams of H2O?

Limiting Reactant

• In a typical reaction, there is one substance that is generally used up first. This reactant is termed as the ‘limiting reactant’ since is limits how much product will be formed. The reactants will be in excess.

DEMO: Setting a table for six people using 10 plates, 8 spoon, and 5 forks.

What is limiting and what is in excess? This is the same with a reaction.

Limiting Reactant

DEMO: CD (5-4) NH3 formation

• Method for calculating the limiting reactant– Calculate the amount of product that would be

formed from each reactant.– The one that produces the least amount of

product is the limiting reactant/reagent.

Limiting Reactant

• What is the maximum amount of sulfur dioxide that can be produced from 95.6 grams of carbon disulfide and 110 grams of oxygen?

CS2 + O2 CO2 + SO2

• What is the maximum amount of HCl that can be produced from 28.2 grams of NaCl, 23.8 grams of H2O, and 32.1 grams of SiO2?

NaCl + H2O + SiO2 HCl + Na2SiO3

Limiting Reactant

• Determine the amount of Cu3P formed if 225 grams of copper is reacted with 65.0 grams of P4. What is the limiting reactant?

– ___Cu + ___P4 ____Cu3P

Determining Percent Yields

• In all the previous calculations, it is assumed that the yield is 100%. In the real world, however, the actual yield is always below 100%. Why?– Many reactions do not go to completion.– Two or more reaction pathways occur

simultaneously.– Some of the product is lost upon isolation.

Determining Percent Yields

• Percent yield

• A 10.0-gram sample of ethanol, C2H5OH, was reacted with excess CH3COOH to produce 14.8 grams of CH3COOC2H5. What is the percent yield?

CH3COOH + C2H5OH CH3COOC2H5 + H2O

%100yieldltheoretica

yieldactualyieldpercenty

Determining Percent Yields

• Aspirin can be synthesized as shown below. For the reaction 12.4 grams of C4H6O3 is reacted with an excess of C7H6O3. The amount of aspirin, C9H8O4, acquired from the reaction is 26.28 grams. What is the percent yield?

2C7H6O3(s) + C4H6O3(l) 2C9H8O4(s) + H2O(l)

Concentrations of Solutions

• In many chemical reactions, the reactants are that mixed are present as solutions.

• Solution consists of a substance (solute) dissolved in another substance (solvent).– DEMO: CuSO4(aq) and Ca(NO3)2(aq)– Mix Ca(NO3)2 and Na2(CO3)

• Ca(CO3) is the solid. Write the chemical equation.

• Solutions are _________ mixtures. Mixing occurs at the molecular level.– Aqueous solutions – solvent in ______.

Concentration of Solutions

• Concentration expresses the amount of solute dissolved in a give amount of solvent or solution.

• Percent by mass of solute

• What mass of NaCl is required to prepare 75.0 grams of a 5.00% solution of NaCl?DEMO: Prepare this solution.

solventofmasssoluteofmasssolutionofmass

%100solutionofmasssoluteofmass

solutepercent

Concentrations of Solutions

• A 7.50% solution of C6H12O6 contains 50.0 grams of C6H12O6. What is the mass of the solution?

• Calculate the mass of NaOH in 150.0 mL of an 8.00% solution of NaOH. The density of the solution is 1.09 g/mL.

Concentrations of Solutions• Molarity is defined as the number of moles of solute

per liter of solution.

DEMO: 1.0 and 0.1 M solution of CuSO4 How many grams of CuSO4 required to make 1 liter of these solutions?

• Prepare 50.0 mL of a 0.100 M solution of aqueous K2Cr2O7.

• What is the molarity of a solution that contains 4.91 grams of HNO3 in 500.0 milliliters of solution?

solutionoflitersofnumbersoluteofmolesofnumber

molarity

Concentration of Solutions

• Determine the mass of BaSO4 required to produce 250.0 milliliters of a 0.125 M solution of barium sulfate.

• Optional: The density of a concentrated HCl solution is 1.185 g/mL and it is 36.31% by mass. What is the molarity?

Solution by Dilution

• When a specified amount of concentrated solution is diluted, the molarity decreases and the volume increases. The number of moles, however, remains constant.– Concentrated solution

• Number of moles = M1 V1

– Diluted solution• Same number of moles = M2 V2

Therefore, M1 V1 = M2 V2

Solutions by Dilution

• M1 V1 = M2 V2 This relationship is used to calculate the amount of concentrated solution, V1, needed to prepare a diluted solution of given volume, V2, and molarity, M2.

• How would you prepare 25.0 mL of a 8.00 10-3 M solution of K2Cr2O7 from 0.100 M K2Cr2O7?

• How many mL of 8.0 M sulfuric acid is required to produce 50.0 mL of a 0.150 M solution of sulfuric acid?

Solutions in Chemical Reactions

• In many chemical reaction it is beneficial or even necessary to add reactants in solution. If molarity is known, the amount of solute (or reactant) to be added to a reaction can be determined.

• What volume of 0.500 M BaCl2 is required to react completely with 4.32 grams of Na2SO4?

Solutions in Chemical Reactions

• What volume of 0.200 M NaOH will react with 50.0 mL of 0.200 M aluminum nitrate? What will be the mass of the aluminum hydroxide precipitate?