Section 9-2:Section 9-2:Ideal Stoichiometric Ideal Stoichiometric

CalculationsCalculations

Coach KelsoeCoach Kelsoe

ChemistryChemistry

Pages 304–311Pages 304–311

Section 9-2 ObjectivesSection 9-2 Objectives

Calculate the amount in moles of a reactant or Calculate the amount in moles of a reactant or product from the amount in moles of a different product from the amount in moles of a different reactant or product.reactant or product.

Calculate the mass of a reactant or product from Calculate the mass of a reactant or product from the amount in moles of a different reactant or the amount in moles of a different reactant or product.product.

Calculate the amount in moles of a reactant or Calculate the amount in moles of a reactant or product from the mass of a different reactant or product from the mass of a different reactant or product.product.

Calculate the mass of a reactant or product from Calculate the mass of a reactant or product from the mass of a different reactant or product.the mass of a different reactant or product.

Ideal Stoichiometric CalculationsIdeal Stoichiometric Calculations

The chemical equation plays a very important The chemical equation plays a very important part in all stoichiometric calculations because part in all stoichiometric calculations because the mole ratio is obtained directly from it. the mole ratio is obtained directly from it. Solving any reaction-stoichiometry problem Solving any reaction-stoichiometry problem must begin with a balanced equation.must begin with a balanced equation.

Chemical equations help us make predictions Chemical equations help us make predictions about chemical reactions without having to run about chemical reactions without having to run the reactions in the laboratory.the reactions in the laboratory.

The calculations we make in this section are The calculations we make in this section are THEORETICAL. They tell us the amounts of THEORETICAL. They tell us the amounts of reactants and products under ideal conditions.reactants and products under ideal conditions.

Ideal Stoichiometric CalculationsIdeal Stoichiometric Calculations

Ideal conditions are rarely met in the lab or Ideal conditions are rarely met in the lab or in industry. Yet, theoretical stoichiometric in industry. Yet, theoretical stoichiometric calculations serve the very important calculations serve the very important function of showing the maximum amount function of showing the maximum amount of product that could be obtained before a of product that could be obtained before a reaction is run in the laboratory.reaction is run in the laboratory.

Solving stoichiometric problems requires Solving stoichiometric problems requires practice. Use a logical and systematic practice. Use a logical and systematic approach along with your previous approach along with your previous knowledge of chemistry to work through knowledge of chemistry to work through these problems.these problems.

Conversions of Quantities in MolesConversions of Quantities in Moles

In these stoichiometric problems, you are In these stoichiometric problems, you are asked to calculate the amount in moles of asked to calculate the amount in moles of one substance that will react with or be one substance that will react with or be produced from the produced from the givengiven amount in moles amount in moles of another substance.of another substance.

The plan for a simple mole conversion The plan for a simple mole conversion problem is:problem is:Amount of Amount of Amount of Amount ofgivengiven unknownunknownsubstance (in mol)substance (in mol) substance (in substance (in mol)mol)

Conversions of Quantities in MolesConversions of Quantities in Moles

These type of problems require only one These type of problems require only one conversion factor: the mole ratio of the conversion factor: the mole ratio of the unknownunknown substance to the substance to the givengiven substance from the balanced equation.substance from the balanced equation.

givengiven quantity x conversion factor = quantity x conversion factor = unknownunknown quantityquantity

Sample Problem 9-1Sample Problem 9-1

In a spacecraft, the carbon dioxide In a spacecraft, the carbon dioxide exhaled by astronauts can be removed exhaled by astronauts can be removed by its reaction with lithium hydroxide, by its reaction with lithium hydroxide, LiOH, according to the following LiOH, according to the following chemical equation:chemical equation: CO CO22(g) + 2LiOH(s) (g) + 2LiOH(s) Li Li22COCO33(s) + H(s) + H22O(l)O(l)

How many moles of lithium hydroxide How many moles of lithium hydroxide are required to react with 20 mol of COare required to react with 20 mol of CO22, , the average amount exhaled by a the average amount exhaled by a person each day?person each day?

Sample Problem 9-1Sample Problem 9-1

Given: amount of COGiven: amount of CO22 = 20 mol; formula- = 20 mol; formula- CO CO22(g) + 2LiOH(s) (g) + 2LiOH(s) Li Li22COCO33(s) + H(s) + H22O(l)O(l)

Unknown: amount of LiOH in molesUnknown: amount of LiOH in moles Answer:Answer:

20 mol CO20 mol CO22 x 2 mol LiOH = 40 mol LiOH x 2 mol LiOH = 40 mol LiOH1 mol CO1 mol CO22

Sample ProblemSample Problem

Ammonia, NHAmmonia, NH33, is widely used as a , is widely used as a fertilizer and in many household fertilizer and in many household cleaners. How many moles of ammonia cleaners. How many moles of ammonia are produced when 6 mol of hydrogen are produced when 6 mol of hydrogen gas react with an excess of nitrogen gas react with an excess of nitrogen gas?gas?

Given: moles of hydrogen gas, HGiven: moles of hydrogen gas, H22,= 6 ,= 6 molmol

Unknown: moles of ammonia, formulaUnknown: moles of ammonia, formula

Sample AnswerSample Answer

NN22 + H + H22 NH NH33 (unbalanced) (unbalanced) NN22 + 3H + 3H22 2NH 2NH33 (balanced) (balanced) 6.0 mol H6.0 mol H22 x 2 mol NH x 2 mol NH33 = 4.0 mol NH = 4.0 mol NH33

3 mol H3 mol H22

Conversions of Amounts in Moles to Conversions of Amounts in Moles to MassMass

In the following calculations, you will be In the following calculations, you will be asked to calculate the mass of a asked to calculate the mass of a substance that will react with or be substance that will react with or be produced from a produced from a givengiven amount in moles amount in moles of a second substance.of a second substance.

This plan requires two conversion This plan requires two conversion factors: mole ratio of the factors: mole ratio of the unknownunknown and and givengiven substances and the molar mass of substances and the molar mass of the the unknownunknown substance. substance.

Conversions of Amounts in Moles to Conversions of Amounts in Moles to MassMass

Amount ofAmount of Amount ofAmount of Mass of Mass ofgivengiven unknownunknown unknownunknownsubstancesubstance substancesubstance substancesubstance(in mol)(in mol) (in mol) (in mol) (in grams) (in grams)

Sample Problem 9-2Sample Problem 9-2

In photosynthesis, plants use energy In photosynthesis, plants use energy from the sun to produce glucose, from the sun to produce glucose, CC66HH1212OO66, and oxygen from the reaction , and oxygen from the reaction of carbon dioxide and water. What of carbon dioxide and water. What mass, in grams, of glucose is produced mass, in grams, of glucose is produced when 3.00 mol of water react with when 3.00 mol of water react with carbon dioxide?carbon dioxide?

Given: moles of water, HGiven: moles of water, H22O, = 3.00 molO, = 3.00 mol Unknown: balanced chemical equation, Unknown: balanced chemical equation,

mass of glucosemass of glucose

Sample Problem 9-2Sample Problem 9-2

COCO22 + H + H22O O C C66HH1212OO66 + O + O22 (unbalanced) (unbalanced) 6CO6CO22 + 6H + 6H22O O C C66HH1212OO66 + 6O + 6O2 2 (balanced)(balanced) 3.00 mol H3.00 mol H22O x 1 mol CO x 1 mol C66HH1212OO66 x 180.18 g C x 180.18 g C66HH1212OO66 = 90.1 g = 90.1 g

6 mol H6 mol H22OO 6 mol 6 mol CC66HH1212OO66

Sample ProblemSample Problem

What mass of carbon dioxide, in grams, What mass of carbon dioxide, in grams, is needed to react with 3.00 mol of His needed to react with 3.00 mol of H22O O in the photosynthetic reaction described in the photosynthetic reaction described in the last problem?in the last problem?

Given: moles of water, HGiven: moles of water, H22O, = 3.00 molO, = 3.00 mol Unknown: balanced chemical equation, Unknown: balanced chemical equation,

mass of carbon dioxidemass of carbon dioxide

Sample ProblemSample Problem

COCO22 + H + H22O O C C66HH1212OO66 + O + O22 (unbalanced) (unbalanced) 6CO6CO22 + 6H + 6H22O O C C66HH1212OO66 + 6O + 6O2 2 (balanced)(balanced) 3.00 mol H3.00 mol H22O x 6 mol COO x 6 mol CO2 2 x 44.01 g CO x 44.01 g CO22 = 132 g = 132 g

6 mol H6 mol H22OO 1 mol CO1 mol CO22

Conversions of Mass to Amounts in Conversions of Mass to Amounts in MolesMoles

In these problems, you start with a In these problems, you start with a mass and converting that mass to mass and converting that mass to moles, then using another conversion moles, then using another conversion factor to convert from your factor to convert from your givengiven substance to your substance to your unknownunknown substance. substance.

Mass ofMass of Amount ofAmount of Amount Amount ofofgivengiven givengiven unknownunknownsubstancesubstance substancesubstance substancesubstance(in grams)(in grams) (in mol) (in mol) (in mol)(in mol)

Sample Problem 9-3Sample Problem 9-3

The first step in the industrial The first step in the industrial manufacture of nitric acid is the catalytic manufacture of nitric acid is the catalytic oxidation of ammonia.oxidation of ammonia.

NH NH33(g) + O(g) + O22(g) (g) NO(g) + H NO(g) + H22O(g)O(g)The reaction is run using 824 g of NHThe reaction is run using 824 g of NH33 and excess oxygen. How many moles of and excess oxygen. How many moles of HH22O are formed?O are formed?

Given: mass of ammonia, NHGiven: mass of ammonia, NH33 = 824 g = 824 g Unknown: balanced chemical equation, Unknown: balanced chemical equation,

moles of Hmoles of H22OO

Sample Problem 9-3Sample Problem 9-3

NHNH33 + O + O22 NO + H NO + H22O (unbalanced)O (unbalanced) 4NH4NH33 + 5O + 5O22 4NO + 6H 4NO + 6H22O (balanced)O (balanced) 824 g NH824 g NH33 x 1 mol NH x 1 mol NH3 3 x 6 mol Hx 6 mol H22O = 72.6 molO = 72.6 mol

17.04 g 17.04 g NHNH33

4 mol 4 mol NHNH33

Mass-Mass CalculationsMass-Mass Calculations

Mass-mass calculations are more practical Mass-mass calculations are more practical than other mole calculations we’ve than other mole calculations we’ve studied.studied.

You can never measure moles directly. You can never measure moles directly. You are generally required to calculate the You are generally required to calculate the amount in moles of a substance from its amount in moles of a substance from its mass, which you can measure in the lab.mass, which you can measure in the lab.

Mass-mass problems can be viewed as the Mass-mass problems can be viewed as the combination of the other types of combination of the other types of problems.problems.

Mass-Mass CalculationsMass-Mass Calculations

Mass ofMass of Amount of Amount of Amount ofAmount of Mass Mass ofofgivengiven given given unknown unknown unknown unknownsubstancesubstance substance substance substance substance substancesubstance(in grams) (in mol)(in grams) (in mol) (in mol) (in mol) (in (in grams)grams)

Sample Problem 9-4Sample Problem 9-4

Tin(II) fluoride, SnFTin(II) fluoride, SnF22, is used in some , is used in some toothpastes. It is made by the reaction of toothpastes. It is made by the reaction of tin with hydrogen fluoride according to tin with hydrogen fluoride according to the following equation.the following equation.

Sn(s) + 2HF Sn(s) + 2HF SnF SnF22(s) + H(s) + H22(g)(g)How many grams of SnFHow many grams of SnF22 are produced are produced from the reaction of 30.00 g of HF with from the reaction of 30.00 g of HF with Sn?Sn?

Given: mass of HF, balanced equationGiven: mass of HF, balanced equation Unknown: mass of SnFUnknown: mass of SnF22

Sample Problem 9-4Sample Problem 9-4 Step 1: Convert grams of given to moles of given.Step 1: Convert grams of given to moles of given. Step 2: Convert moles of given to moles of unknown.Step 2: Convert moles of given to moles of unknown. Step 3: Convert moles of unknown to grams of unknown.Step 3: Convert moles of unknown to grams of unknown. 30.00 g HF x 1 mol HF x 1 mol SnF30.00 g HF x 1 mol HF x 1 mol SnF22 x 156.71 g SnF x 156.71 g SnF22 = 117.5 g = 117.5 g

20.01 g 20.01 g HFHF

2 mol HF2 mol HF 1 mol 1 mol SnFSnF22