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Chapter 8Quantities in

Chemical Reactions

Chapter 8Quantities in

Chemical Reactions

Tro, 2nd ed.

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STOICHIOMETRY: SOME REVIEW

The molar mass of an element is its atomic mass in grams.

It contains 6.0221 x 1023 atoms (Avogadro’s number) of the element.

The molar mass of an element or compound is the sum of the atomic masses of all its atoms.

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Avogadro’s Number of Particles

6.0221 x 1023

Particles

Molar Mass

1 MOLE

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1 mole NaClmoles NaCl = 292.15 grams NaCl = 5.000 moles NaCl

58.443 g NaCl

1 molemoles = grams

molar mass

How many moles of NaCl are present in 292.215 grams of NaCl? The molar mass of NaCl =58.443 g.

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STOICHIOMETRY

Consider the combustion of acetylene: 2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l) 2 molecules 5 molecules 4 molecules 2 molecules200 “ 500 “ 400 “ 200 “12x1023 31x1023 24x1023 12x1023

2 moles 5 moles 4 moles 2 moles Consider each of the formulas as molecules or molesCoefficients represent number of moles“Recipe” just like a cooking recipe:

2 cups of acetylene requires 5 cups of oxygen, etc. When we use the coefficients to relate the chemicals in an equation,

we are doing STOICHIOMETRY

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STOICHIOMETRY

2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l)

EXAMPLE: If 6 moles of C2H2 are burned, then 15 moles of O2 are needed, 12 moles of CO2 are made and 6 moles of H2O are made

How did we know? We used ratios.

Practice with 0.102 moles of C2H2:

0.102 mol * 5 mol O2/2 mol C2H2 = 0.255 mol O2

0.102 mol * 4 mol CO2/2 mol C2H2 = 0.204 mol CO2

0.102 mol * 2 mol H2O/2 mol C2H2 = 0.102 mol H2O

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Introduction to Stoichiometry:The Mole-Ratio Method

Stoichiometry: The area of chemistry that deals with the quantitative relationships between reactants and products.

Mole Ratio: a ratio between the moles of any two substances involved in a chemical reaction.

The coefficients used in mole ratio expressions are derived from the coefficients used in the balanced equation.

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2

23 m

1 mol

ol H

N

N2 + 3H2 2NH31 mol 2 mol3 mol

The Mole-Ratio Method

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1 mol 2 mol3 molN2 + 3H2 2NH3

2

32 mo

3 mol H

l NH

The Mole-Ratio Method

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The Mole-Ratio Method

The mole ratio is used to convert the number of moles of one substance to the corresponding number of moles of another substance in a stoichiometry problem.

The mole ratio is used in the solution of every type of stoichiometry problem.

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The Mole Ratio Method for Stoichiometry

1. Have a balanced chemical equation.

2. Convert the quantity of starting substance to moles (if it is not already moles)

3. Convert the moles of starting substance to moles of desired substance.

4. Convert the moles of desired substance to the units specified in the problem.

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moles of desired substance in the equationmoles of desired substance = moles of starting substance

moles of starting substance in the equation

In the following reaction how many moles of PbCl2

are formed if 5.000 moles of NaCl react?2NaCl(aq) + Pb(NO3)2(aq) PbCl2(s) + 2NaNO3(aq)

5.000 moles NaCl 2moles of PbCl = 22.500 mol PbCl21 mol PbCl

2 mol NaCl

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Mole Ratio

Calculate the number of moles of phosphoric acid (H3PO4) formed by the reaction of 10 moles of sulfuric acid (H2SO4).

Ca5(PO4)3F + 5H2SO4 3H3PO4 + HF + 5CaSO4

Step 2 Moles starting substance: 10.0 mol H2SO4

Step 3 The conversion needed is moles H2SO4 moles H3PO4

1 mol 5 mol 3 mol 1 mol 5 mol

3 42 4

2 4

3 mol H PO10 mol H SO x =

5 mol H SO 3 46 mol H PO

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Mole-Mass Calculations

The object of this type of problem is to calculate the mass of one substance that reacts with or is produced from a given number of moles of another substance in a chemical reaction.

If the mass of the starting substance is given, we need to convert it to moles. (Step 2)

We use the mole ratio to convert moles of starting substance to moles of desired substance. (Step 3)

We can then change moles of desired substance to mass of desired substance if called for by the problem. (Step 4)

15Mole Ratio

2 4

3 4

5 mol H SO =

3 mol H PO

Calculate the number of moles of H2SO4 necessary to yield 784 g of H3PO4.

Ca5(PO4)3F+ 5H2SO4 3H3PO4 + HF + 5CaSO4

Method 1 Step by Step Calculations

Step 1 Balance reaction given, with 784 grams of H3PO4.

Step 2 Convert grams of H3PO4 to moles of H3PO4.

Step 3 Convert moles of H3PO4 to moles of H2SO4 by the mole-ratio method.

3 4

3 4

1 mol H PO =

98.0 g H PO

3 48.00 mol H PO

2 413.3 mol H SO

3 4784 g H PO

3 48.00 mol H PO

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Mole Ratio

2 4

3 4

5 mol H SO =

3mol H PO

Calculate the number of moles of H2SO4 necessary to yield 784 g of H3PO4

Ca5(PO4)3F+ 5H2SO4 3H3PO4 + HF + 5CaSO4

Method 2 Continuous Calculation

grams H3PO4 moles H3PO4 moles H2SO4

The conversion needed is

3 4784 g H PO 3 4

3 4

1 mol H PO

98.0 g H PO

2 413.3 mol H SO

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Mass-Mass Calculations

Solving mass-mass stoichiometry problems requires all four steps of the mole-ratio method.

You list the four steps here:

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Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2.

N2 + 3 H2 2 NH3Method 1 Step by Step Calculations

Step 2 Convert 112 g of H2 to moles.

grams moles

2

2

1 mol H

2.02 g H

2112 g H 255.4 moles H

Step 3 Calculate the moles of NH3 by the mole ratio method.

3

2

2 mol NH=

3 mol H

255.4 moles H 336.9 moles NH

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Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2.

N2 + 3 H2 2 NH3

Step 4 Convert moles NH3 to grams NH3.

moles grams

336.9 moles NH 3

3

17.0 g NH=

1 mol NH

3629 g NH

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Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2.

N2 + 3 H2 2 NH3

grams H2 moles H2 moles NH3 grams NH3

2

2

1 mol H

2.02 g H

2112 g H 3

2

2 mol NH

3 mol H

Method 2 Continuous Calculation

3

3

17.0 g NH=

1 mol NH

3629 g NH

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GROUP PRACTICE:

First write the balanced chemical reaction: ammonium phosphate reacts with calcium hydroxide to produce calcium phosphate, ammonia and water. Then answer:

a. What weight of calcium hydroxide is need to produce 155 g of calcium phosphate?

b. How many molecules of ammonia gas are released? c. What volume of water will be produced, assuming

normal conditions?

2 (NH4)3PO4(aq) + 3 Ca(OH)2(aq) Ca3(PO4)2(aq) + 6 NH3(g) + 6 H2O(l)

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a. 155 g Ca3(PO4)2 * 1mol/310 g = 0.500 mol Ca3(PO4)2

0.500 mol Ca3(PO4)2 * 3Ca(OH)2/1 Ca3(PO4)2 = 1.50 mol Ca(OH)2

1.50 mol Ca(OH)2 * 74 g/mol = 111 g

b. 0.5 mol Ca3(PO4)2* 6 H2O/1 Ca3(PO4)2 = 3.0 mol H2O

3.0 mol H2O * 18 g/mol * 1 mL/1 g = 54 mL water

c. 0.5 mol Ca3(PO4)2 * 6 NH3/1 Ca3(PO4)2 = 3.0 mol NH3

3.0 mol NH3 * 6.0221 x 1023 molecules/mol = 1.8 x 1024 molecules

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Limiting-Reactant and Yield Calculations

The limiting reactant is one of the reactants in a chemical reaction.

It is called the limiting reactant because the amount of it present is insufficient to react with the amounts of other reactants that are present.

The limiting reactant limits the amount of product that can be formed.

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H2 + Cl2 2HCl

+

7 molecules H2 can form 14 molecules HCl

4 molecules Cl2 can form 8 molecules HCl 3 molecules of H2 remain

H2 is in excess Cl2 is the limiting

reactant 9.3

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Steps Used to Determine the Limiting Reactant

1. Calculate the amount of product (moles or grams, as needed) formed from each reactant.

2. Determine which reactant is limiting. The reactant that gives the least amount of product is the limiting reactant; the other reactant is in excess.

3. Calculate the amount of the other reactant required to react with the limiting reactant, then subtract this amount from the starting quantity of the reactant. This gives the amount of the that substance that remains unreacted.

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How many moles of HCl can be produced by reacting 4.0 mol H2 and 3.5 mol Cl2? Which compound is the limiting reactant?

Step 1 Calculate the moles of HCl that can form from each reactant.

24.0 mol H2

2 mol HCl

1 mol H

8.0 mol HCl

23.5 mol Cl2

2 mol HCl

1 mol Cl

7.0 mol HCl

H2 + Cl2 → 2 HCl

Step 2 Determine the limiting reactant.

The limiting reactant is Cl2 because it produces less HCl than H2.

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You Practice:

Work on Skillbuilders 8.4 & 8.5 with a partner. Turn in one piece of paper with both your names on it showing all work.

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How many moles of silver bromide (AgBr) can be formed when solutions containing 50.0 g of MgBr2 and 100.0 g of AgNO3 are mixed together? How many grams of the excess reactant remain unreacted?

Step 1 Calculate the grams of AgBr that can form from each reactant.

MgBr2(aq) + 2 AgNO3 (aq) → 2 AgBr(s) + Mg(NO3)2(aq)

The conversion needed isg reactant → mol reactant → mol AgBr → g AgBr 250.0 g MgBr 102 g AgBr2

2

1 mol MgBr

184.1 g MgBr

2

2 mol AgBr

1 mol MgBr

187.8 g AgBr

1 mol AgBr

3100.0 g AgNO 110.5 g AgBr3

3

1 mol AgNO

169.9 g AgNO

3

2 mol AgBr

2 mol AgNO

187.8 g AgBr

1 mol AgBr

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How many grams of the excess reactant (AgNO3) remain unreacted?

Step 3 Calculate the grams of unreacted AgNO3. First calculate the number of grams of AgNO3 that will react with 50 g of MgBr2.

250.0 g MgBr 392.3 g AgNO2

2

1 mol MgBr

184.1 g MgBr

3

2

2 mol AgNO

1 mol MgBr

3

3

169.9 g AgNO

1 mol AgNO

The conversion needed isg MgBr2 → mol MgBr2 → mol AgNO3 → g AgNO3

The amount of AgNO3 that remains is

100.0 g AgNO3 - 92.3 g AgNO3 = 7.7 g AgNO3

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Theoretical Yield

The quantities of products calculated from equations represent the maximum yield (100%) of product according to the reaction represented by the equation.

Many reactions fail to give a 100% yield of product.

This occurs because of side reactions and the fact that many reactions are reversible.

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Theoretical Yield & Percent Yield

The theoretical yield of a reaction is the calculated amount of product that can be obtained from a given amount of reactant.

The actual yield is the amount of product finally obtained from a given amount of reactant.

The percent yield of a reaction is the ratio of the actual yield to the theoretical yield multiplied by 100.

Actual yield * 100 = Percent Yield Theor yield

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187.8 g AgBr

1 mol AgBr

Silver bromide was prepared by reacting 200.0 g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if 375.0 g of silver bromide was obtained from the reaction:

MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)

Step 1 Determine the theoretical yield by calculating the grams of AgBr that can be formed.

The conversion needed isg MgBr2 → mol MgBr2 → mol AgBr → g AgBr

2200.0 g MgBr 408.0 g AgBr2

2

1 mol MgBr

184.1 g MgBr

2

2 mol AgBr

1 mol MgBr

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Silver bromide was prepared by reacting 200.0 g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if 375.0 g of silver bromide was obtained from the reaction:

MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)

Step 2 Calculate the percent yield.

actual yieldpercent yield = x 100

theoretical yield

percent yield = 375.0 g AgBr

x 100 =408.0 g AgBr

91.9%

must have same units

must have same units

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You Practice:

Work on Skillbuilder 8.6 and problem 60 on page 257.