Reactions in ChemistryBalancing Equations

Balanced equations•Balanced equations reflect the fact that

matter is neither created nor destroyed in a chemical reaction.

•An equation is balanced if the same number of each type of atom is found in both the reactants and products of the reaction.

•A total count must be taken on each side and the numbers compared to determine whether or not an equation is balanced.

Counting reactants•There are two reactants in the following

reaction. Determine the number of atoms of each type that are listed on the reactant side:

CaBr2: AgNO3:

Ca – Ag – Br – N –

O –

2332 )Ca(NOAgBr2AgNO CaBr

Counting products•There are two products in this reaction.

Determine the number of atoms of each type that are listed on the product side:

AgBr: Ca(NO3)2:

Ag – Ca – Br – N –

O –

2332 )Ca(NOAgBr2AgNO CaBr

Counting products•As you can see below, in a properly balanced

equation there will be the same number of each type of atom in the reactants and products:

REACTANTS: PRODUCTS:CaBr2: AgNO3: AgBr: Ca(NO3)2

Ca – 1 Ag – 2 Ag – 2 Ca – 1Br – 2 N – 2 Br – 2 N – 2

O – 6 O – 6

2332 )Ca(NOAgBr22AgNO CaBr

Counting reactants•Notice that in the following reaction oxygen

shows up in both of the reactants; the total count of reactant atoms, once these are combined, is:

FeSO4: Ba(NO3)2 All reactant atoms:

Fe – 1 Ba – 1 Fe – Ba – S – 1 N – 2 S – N – O – 4 O – 6 O –

234234 )Fe(NOBaSO)Ba(NOFeSO

Counting reactants•Remember this as you count the atoms

on the product side as well:

BaSO4: Fe(NO3)2 All product atoms:

Ba – Fe – Fe – Ba – S – N – S – N – O – O – O –

234234 )Fe(NOBaSO)Ba(NOFeSO

Comparing the counts•As we compare the counts of reactant and

product atoms, we see that they are the same:

All reactant atoms: All product atoms:Fe – 1 Ba – 1 Fe – 1 Ba – 1S – 1 N – 2 S – 1 N – 2O – 10 O – 10

234234 )Fe(NOBaSO)Ba(NOFeSO

Balancing equations•What if you are given a chemical equation

showing the reactants and products and are asked to balance it?

• In order to balance the equation, we must insert coefficients that will make the count of atoms the same on the two sides of the reaction.

•We CANNOT change the formulas themselves, because then we would be describing a different reaction taking place between different chemicals.

Why can’t we change formulas?•What if one of the

reactants was water (H2O) and we needed another oxygen atom to balance the equation?

•If we changed the formula so that it had 2 atoms of oxygen, we wouldn’t have water any more! wps.prenhall.co

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Balancing equations•First, take an atom inventory.

•Pick an element to work with and see what coefficient could be inserted in front of a formula in order to make that element have the same number on both sides.

•Redo your inventory and see how the atoms of reactants and products compare now.

•Keep doing this until balanced.

•Make sure the final coefficient ratio is simplified.

Take an atom inventory•Balance the following equation:

Reactants ProductsBr – Br – Na – Na – I – I –

22 INaBrNaIBr

•Note that there are different amounts of bromine and iodine atoms on each side.

Pick an element to work with•Let’s choose bromine. We need to get 2

bromine atoms on the product side. We CANNOT change the formula. Try putting a coefficient in front of the formula that contains the bromine atoms.

22 INaBr2NaIBr 22 INaBrNaIBr

•How does this affect the count?

Redo your inventory

Reactants ProductsBr – Br – Na – Na – I – I –

•Now our sodium atoms don’t match. How can we get them to be the same?

22 INaBr2NaIBr

Pick another element to work with•To get the sodium counts to match, we need

to put a coefficient in front of the compound containing sodium on the reactant side.

22 INaBr2NaI2Br

Reactants ProductsBr – Br – Na – Na – I – I –

22 INaBr2NaIBr

Check for simplified ratio•Now that all of the counts match, showing

that there are the same number of each type of atom on each side (no matter is being created or destroyed), we need to check the coefficient ratio and see if it can be simplified:

•The ratio is 1:2:2:1.

•These numbers have no common factors, so the ratio is in simplified form.

22 INaBr2NaI2Br

Check for simplified ratio•Why must we perform this step?

•The equation is also balanced if written as:

However, notice that the coefficient ratio is 2:4:4:2, all of which are divisible by 2, so it would reduce to 1:2:2:1. We always write equations with the smallest coefficients that will balance the equation.

22 I2NaBr4NaI42Br

Try another example•Now try balancing the following equation,

remembering to check your full atom count each time you insert a new coefficient:

Reactants ProductsAg – Ag – N – N – O – O – H – H – S – S –

3223 HNOSAgSHAgNO

Example 3•Do an initial inventory for the following

equation:

Reactants ProductsAl – Al – Cl – Cl –

32 AlClClAl

•If coefficients must be whole numbers, how do we get the number of chlorine atoms to match?

Example 3•What we need is a number that is a

multiple of both 2 and 3. The least common multiple of 2 and 3 is 6.

•How do we get 6 chlorine atoms on each side?

Reactants ProductsAl – Al – Cl – Cl –

32 AlClClAl

Example 4•Now try balancing the following equation,

remembering to check your full atom count each time you insert a new coefficient:

Reactants ProductsFe – Fe – O – O –

322 OFeOFe

Challenge Problem #1•Now try balancing the following equation.

(HINT: If a polyatomic ion appears unchanged on both sides of the equation, you can treat it as a single unit.)

Reactants ProductsAl – Al – SO4 – SO4 –

Ca – Ca – OH – OH –

432342 CaSOAl(OH)Ca(OH))(SOAl

Challenge Problem #2•Now try balancing the following equation.

(HINT: Begin by balancing elements that appear only once on each side of the equation.)

Reactants ProductsFe – Fe – O – O – H – H –

OHOFeFe(OH) 2323

Challenge Problem #3•Now try balancing the following equation.

Reactants ProductsC – C – H – H – O – O –

OHCOOOHC 222112212

•Not all equations end up with small coefficients!

Balancing summary•DO NOT change formulas, only coefficients!

•Redo inventory each time a coefficient is changed.

• It is generally safe to leave polyatomics together if they appear in both the reactants and products.

•Begin balancing elements that only appear once on each side of the equation.

•Be sure the final coefficient ratio is simplified.

Resources•http://

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