Interpreting Chemical Equations - Arabia Mountain High School

[Pages:6]Interpreting Chemical Equations

Goal: Given a balanced chemical equation or information from which a chemical equation can be written, describe its meaning at the particulate, molar and macroscopic levels.

Example 1:

2H2(g)

+ O2(g)

2H2O(g)

Particulate Level: 2 molecules 1 molecule

2 molecules

Molar Level:

2 moles

1 mole

2 moles

Macroscopic Level: 4.04 g

32.00g

36.04 g

Example 2:

2C2H6(g) + 7O2(g) 4CO2(g)

+ 6H2O (l)

Particulate Level: 2 molecules 7molecules 4 molecules

6 molecules

Molar Level:

2 moles

7 moles

4 moles

6 moles

Macroscopic Level: 60.16 g

224.00 g

176.04 g

108.12 g

Quantity Relationships in Chemical Reactions, (Stoichiometry)

Goal: Given a chemical equation, or a reaction for which the equation is known, and the number of moles of one species in the reaction, calculate the number of moles of any other species.

Consider the equation:

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

1. How many O2 molecules are required to react with 308 molecules of NH3?

Answer: On the particulate level,

4 molecules of NH3 react with 5 molecules of O2

That is

# $%&'()&'* +, OR - $%&'()&'* ./0

- $%&'()&'* ./0

# $%&'()&'* +,

Recall that: Required number and units = Given number and units x conversion factor

Therefore,

# O2 molecules = 308 NH3 molecules

x

+, ./0

= 385 O2 molecules

2. If 3.20 moles of NH3 react according to the above equation, how many moles of H2O will be

produced?

Answer: The unit path is mol NH3 mol H2O

Therefore, use : $%& /,+ as the conversion factor

- $%& ./0

Hence, mole H2O = 3.20 mol NH3

x

: $%& /,+ - $%& ./0

= 4.80 mol H2O

3. How many moles of oxygen are required to burn 2.40 moles of ethane, C2H6?

Answer: First, write a balanced equation for the reaction described. Hint: Remember that combustion of hydrocarbon always forms CO2 and H2O

2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O (l)

The conversion factor from the relationship between the given and the wanted quantity using the coefficients in the equation is ; $%& +,

< $%& =,/>

Hence, mol O2 = 2.40 mol C2H6 x

; $%& +, < $%& =,/>

=

8.40

mol

O2

Mass Calculations Mass calculations may involve: a. Writing a balanced equation b. Calculating molar masses from chemical formulas c. Using molar masses to change from mass to moles or from moles to mass d. Using the equation to change from moles of one species to moles of another

The mass ?to- mass path is:

Mass of given moles of given moles of wanted mass of wanted

Mass given x $%& ?@A'B x $%& DCBE'F x $C** DCBE'F

$C** ?@A'B $%& ?@A'B

$%& DCBE'F

Example: Calculate the number of grams of oxygen that are required to burn 155 g of ethane in the reaction: 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O (l)

Answer:

Number of gram O2 = 155 g C2H6 x

G $%& =, /> x

HI.I; ? =, />

; $%& +, < $%& =, />

x H

x

H

GII.

x

; $%& =+, G $%& =K/L>

x

--.IG ? =+, G $%& =+,

= 203 g CO2

Limiting and Excess Reactants, Theoretical Yield, Actual Yield and Percent Yield

Definitions:

Limiting reactant: the reactant that is completely used up during the reaction is called the limiting reactant.

Excess reactant: the reactant that some of it remains unreacted after the reaction is complete.

Theoretical Yield: the amount of product formed from the complete conversion of the given amount of reactant to product. Theoretical yield is always a calculated quantity based on the stoichiometry, (ratio) of the reaction equation.

Actual Yield: a measured quantity determined by experiment. Factors such as impure reactants, incomplete reaction and side reactions cause the actual yield to be less than the theoretical yield.

Percent Yield: this is the actual yield expressed as a percentage of the theoretical yield. % Yield = C(E)C& N@'&F x 100

EO'%P'E@(C& N@'&F

Understanding Limiting reactant

a. How many pairs of gloves can you assemble from 20 left gloves and 30 right gloves? Answer: 20 pairs theoretical yield

b. How many unmatched gloves will be left over? Answer: 10 right gloves excess amount

c. Which hand will these unmatched gloves fit? Answer: Right excess reactant

d. What prevented you from assembling more than 20 pairs? Answer: Left gloves limiting reactant

How To Solve Limiting Reactant Problems There are two methods:

a. Smaller-Amount method b. Comparison-of-Moles Method

We shall use the smaller-amount method

Procedure:

1. Calculate the amount of product that can be formed by the initial amount of each reactant a. The reactant that yields the smaller amount of product is the limiting reactant.

b. The smaller amount of product is the amount (theoretical yield) that will be formed when all of the limiting reactant is used up.

2. Calculate the amount of excess reactant that is used by the total amount of the limiting reactant.

3. Subtract from the amount of excess reactant present initially, the amount that is used by the limiting reactant. The difference is the amount of excess reactant that is left.

Example: Calculate the mass of antimony (III) chloride, SbCl3, that can be produced by the reaction of 129 g antimony, Sb, and 106 g chlorine. How much of the excess reactant is left?

Answer:

2Sb +3Cl22SbCl3

Molar mass (g/mol) 121.8 71

228.3

Mass Used (g)

129 106

Moles Used, (mol) G ................
................

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