Calculations From Chemical Equations - Kids in Prison Program · 2015-12-30 · 5 2Al + Fe 2 2 O 3 Fe + Al 2 O 3 •For calculations of mole-mass-volume relationships in a chemical

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Calculations From

Chemical EquationsChapter 9

Hein and Arena

Eugene Passer

Chemistry Department

Bronx Community College

© John Wiley and Sons, Inc.

Version 1.1

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A Short Review

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• The molar mass of an element is its

atomic mass in grams/mol.

• It contains 6.022 x 1023 atoms

(Avogadro’s number) of the element.

• The molar mass of a compound is the

sum of the atomic masses of all its

atoms of each element.

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2 C = 2(12.01 g) = 24.02 g

6 H = 6(1.01 g) = 6.06 g

1 O = 1(16.00 g) = 16.00 g

46.08 g/mol

Calculate the molar mass of C2H6O.

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2 2Al + Fe2O3 Fe + Al2O3

• For calculations of mole-mass-volume

relationships in a chemical equation.

– The chemical equation must be balanced by

using smallest whole number coefficients.

2 mol 2 mol1 mol 1 mol

The equation is balanced.

– The coefficient in front of a formula

represents the number of moles, atoms,

ions or molecules, of the reactant or

product.

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Introduction to Stoichiometry:

The Mole Ratio Method

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• Stoichiometry: The area of chemistry

that deals with the quantitative

relationships between reactants and or

products.

• Mole Ratio: A ratio between the moles

of any two substances involved in a

chemical reaction (Conversion Factor).

– The coefficients used in mole ratio

expressions are derived from the

coefficients used in the balanced

equation.

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Examples

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2

23 m

1 mol

ol H

N

N2 + 3H2 2NH31 mol 2 mol3 mol

10

1 mol 2 mol3 mol

N2 + 3H2 2NH3

2

32 mo

3 molH

l NH

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• The mole ratio is used to convert moles

of one substance to moles of another

substance in a stoichiometry problem.

mol. H2 mol. N2

• The mole ratio is used to solve every type

of stoichiometry problem.

mole ratio

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The Mole Ratio Method: Step by Step

1. Convert the quantity of starting substance

to moles (if it is not already in moles)

2. Convert the moles of starting substance to

moles of desired substance.

3. Convert the moles of desired substance to

the units specified (i.e. grams, molecules,

atoms, etc.) in the problem.

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Mole-Mole

Calculations

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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 1 Moles starting substance: 10.0 mol H2SO4

Step 2 The conversion sequence 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 SO3 46 mol H PO

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Step 2 The conversion sequence is …

moles Ca5(PO4)3F moles H2SO4

Calculate the number of moles of sulfuric acid

(H2SO4) that react when 10 moles of Ca5(PO4)3F react.

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

Mole Ratio

Step 1 The starting substance is 10.0 mol Ca5(PO4)3F

1 mol 5 mol 3 mol 1 mol 5 mol

2 45 4 3

5 4 3

5 mol H SO10 mol Ca (PO ) F x =

1 mol Ca (PO ) F2 450 mol H SO

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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

moles NaCl moles PbCl2

The conversion sequence is …

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Mole-Mass

Calculations

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Examples

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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

grams H3PO4 moles H3PO4 moles H2SO4


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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Mole Ratio

2

3

3 mol H

2 mol NH

moles NH3 moles H2 grams H2

Calculate the number of grams of H2 required to form

12.0 moles of NH3.

N2 + 3H2 2NH3


312.0 mol NH 2

2

2.02 g H =

1mol H

36.4 g H2

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Mass-Mass

Calculations

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Calculate the number of grams of NH3 formed by the

reaction of 112 grams of H2.

N2 + 3H2 2NH3

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

3

3

17.0 g NH=

1 mol NH

628 g NH3

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Limiting-Reactant and

Yield Calculations

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Limiting Reactant

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A limiting reactant is a reactant in a

chemical reaction with insufficient mass

to react with all the mass of the other

reactant(s) present.

The limiting reactant limits the amount

of product that can be formed.

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How many bicycles

can be assembled

from the parts

shown?

From eight wheels four

bikes can be constructed.

From four frames four

bikes can be constructed.

From three pedal assemblies

three bikes can be constructed.

The limiting part is the

number of pedal

assemblies.

9.2

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Steps Used to Determine the Limiting Reactant

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1. Calculate the moles of each reactant(This is what you HAVE).

2. Convert moles of each reactant to moles of

the other reactant (This is what you NEED).

3. Determine limiting reactant by:

HAVE – NEED > 0 ; NOT LIMITING

HAVE – NEED < 0 ; LIMITING

4. Excess reactant is the difference between:

HAVE – NEED of NON- LIMITING reactant.

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Examples

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How many grams of H2 will be produced if 15.0 g of

HCl are reacted with 7.5 g Mg? Assume 100% yield.

Mg + 2HCl MgCl2 + H2

Step 1: Calculate moles of reactants you have.

Step 2: Calculate moles of reactants you need.

Step 3: Determine limiting reactant.

Step 4: Calculate desired quantity based on

limiting reactant.

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Have 0.31 0.41

Need

Moles have (Step 1):

x mol Mg = 7.5g Mg x 1mol/24.31g Mg = 0.31mol. Mg

x mol HCl = 15.0g HCl x 1mol/36.45g HCL = 0.41 mol HCl

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Have 0.31 0.41

Need 0.21 0.62

Moles need (Step 2):

x mol Mg = 0.41mol HCl x 1m Mg/2 m HCl = 0.21mol. Mg

x mol HCl = 0.31 mol Mg x 2 m HCl/1m Mg = 0.62 mol HCl

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Have 0.31 0.41

Need 0.21 0.62

Limiting reactant (Step 3):

The limiting reactant is HCl, since we

have less than we need:

have - need < 0; 0.41 - 0.62 = - 0.21

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The grams of H2 is calculated from the mass of the limiting

reactant HCl.

g H2 15.0g HCl x 1mol. HCl x 1mol. H2 x 2.02g H2 =

36.45g HCl 2 mo HCl 1 mol H2

0.42g H2

Suppose 28.82 grams of carbon and 55.86 grams of steam

react. How many grams of hydrogen will be produced?

Assume 100% yield.

C (s) + H2O (g) CO (g) + H2 (g)

C (s) + H2O (g) CO (g) + H2 (g)

Moles have 2.40 3.10

Moles need 3.10 2.40

0.70 moles “xs” H2O

Therefore, the carbon is the limiting reactant, since we have less carbon than

we need to completely react with all the water we have.

The mass of hydrogen is calculated using the limiting reactant;

28.82 g C x 1 mol C x 1 mol H2 x 2.02 g H2 = 4.847 g H2

12.01 g C 1 mol C 1 mol H2

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Reaction Yield

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The quantities of products calculated

from equations represent the maximum

yield (100%) of product according to the

reaction represented by the equation.

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Many reactions fail to give a 100%

yield of product.

This occurs because of:

side reactions

purification steps

reversible reactions

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• 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 “actually obtained” from a

given amount of reactant.

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• The percent yield of a reaction is the

ratio of the actual yield to the theoretical

yield multiplied by 100.

actual yield x 100 = percent yield

theoretical 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.


g 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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Calculations From Chemical Equations - Kids in Prison Program · 2015-12-30 · 5 2Al + Fe 2 2 O 3 Fe + Al 2 O 3 •For calculations of mole-mass-volume relationships in a chemical