Notes: Chemical Reactions and Balancing Equations

Below is a chemical equation of a chemical reaction between sodium and water.

REACTANTS PRODUCTS

2 Na + 1 I2 ‡ 2 NaI

Coefficients subscripts

Reactants: substances present before a chemical reaction occurs

Products: substances present at the end after a chemical reaction has occurred

* From reactants to products, the chemical reaction allows bonds to breaks apart, rearrange atoms, and forms new bonds to produce “new” substances with different properties.

Coefficient: whole number multipliers that get distributed to each atom in a chemical formula

Subscript: a whole number that follows an atom to indicate more than one is bonded in a chemical formula

A. Balancing Equations:1) List each element found in the equation on the left hand side. Draw a vertical dotted line below

the arrow, where the chemical change occurs2) Count the number of atoms for each element on BOTH sides of the equation3) Use coefficients in front of chemical symbols or formulas to balance the numbers between the

reactants and the products

Ex. #1 ___H2 + ___O2‡ ___H2O _2_H2 + ___O2‡ _2_H2O

H: 2 H: 2 H: 2 4 H: 2 4

O: 2 O: 1 O: 2 O: 1 2

Ex. #2 ___Na + ___H2O ‡ ___NaOH + ___H2 _2_Na + _2_H2O ‡ _2_NaOH + ___H2

Na: 1 Na: 1 Na: 1 2 Na: 1 2

H: 2 H: 3 H: 2 4 H: 3 4

O: 1 O: 1 O: 1 2 O: 1 2

Balancing chemical equations simply takes time and practice to master!!

B. Law of Conservation of Mass: In a balanced chemical reaction, the number, type and amount of atoms that are started with are equal to the number and type of atoms ending with

Ex: A reaction began with 7.45 g of water. After decomposing the water, it was determined that 2.90 grams of hydrogen gas was produced along with oxygen gas. Given the following reaction, determine the amount of O2 created.

2 H2O(l) ‡ 2 H2(g) + O2(g)7.45 g = 2.90 g + X

- 2.90 -2.90

X = 4.55 g

Ex: A reaction began with 20.0 grams of Zinc and 15.0 g of hydrogen chloride. After the reaction occured, it was determined hydrogen gas was produced along with 32.5 grams of zinc chloride. Given the following reaction, determine the amount of H2 created.

Zn(s) + 2 HCl(l) ‡ H2(g) + ZnCl2

20.0 g + 15.0 g = X + 32.5 g- 32.5 -32.5

X = 2.50 g

C. Writing Reactions and Balancing Equations:When writing reactions, be sure to take the time to look up charges on the PT and correctly write each formula (remember: compounds and elements need to have no overall charge)

Example 1: Magnesium and lead IV nitrate react to form magnesium nitrate and lead

___ Mg + ___ Pb(NO3)4‡ ___ Mg(NO3)2 + ___ Pb

_2_ Mg + ___ Pb(NO3)4‡ _2_ Mg(NO3)2 + ___ Pb

Example 2: Copper and oxygen gas react to form copper I oxide

___ Cu + ___ O2‡ ___ Cu2O

_4_ Cu + ___ O2‡ _2_ Cu2O

Example 3: Calcium and water react to form calcium hydroxide and hydrogen gas

___ Ca + ___ H2O ‡ ___ Ca(OH)2 + H2

___ Ca + _2_ H2O ‡ ___ Ca(OH)2 + H2

D. Types of ReactionsThere are 5 basic reaction that all reactions can be classified as. You will need to be able to classify the types of reaction as well as to predict the products (or reactants) when given.

1) Synthesis Reaction - this reaction is also sometimes known as a combination reaction- Two or more single elements combine and react to form only 1 product- General formula: A + B ‡ AB A + B + C ‡ ABC

Example 1: Sodium + Chlorine ‡ _______________

Work: Sodium + Chlorine ‡ Sodium Chloride

Na + Cl2 ‡ NaCl

_2_Na + ___Cl2‡ _2_NaCl

Example 2: _______ + _______ ‡ Iron III oxide

Work: Iron + Oxygen ‡ Iron III oxide

Fe + O2 ‡ Fe2O3

_4_Fe + _3_O2 ‡ _2_Fe2O3

2) Decomposition Reaction - There is only 1 reactant, and it breaks down to form two or more single elements - General formula: AB ‡ A + B ABC ‡ A + B + C

Example 1: Carbon Dioxide‡ _______ _ + _________

Work: Carbon Dioxide‡ Carbon + Oxygen

CO2 ‡ C + O2

_1_CO2 ‡ _1_C + _1_O2

Example 2: __________‡ Copper (I) + Oxygen

Work: Copper I oxide ‡ Copper (I) + Oxygen

Cu2O ‡ Cu + O2

_2_Cu2O ‡ _4_Cu + _1_O2

3) Combustion Reactions- The reactants are ALWAYS a CxHy compound that reacts with O2, and the products are

ALWAYS CO2 and H2O- General Formula: CxHy + O2‡ CO2 + H2O

Example 1: Carbon IV hydride + oxygen ‡ __________ + ___________

CH4 + O2‡ __________ + ___________

CH4 + 2 O2‡ CO2 + 2 H2O

Example 2: Glucose + oxygen ‡ __________ + ___________

C6H12O6 + O2‡ __________ + ___________

C6H12O6 + 6 O2‡ 6 CO2 + 6 H2O

4) Single Replacement Reactions: A reaction in which a single element can replace another element in a compound.

General Formula: A+ + B+C- ‡ A+C- + B+ or A- + B+C- ‡ B+A- + C-

This can occur if:a) The element to be replaced is the same charge as the element trying to replace it

ex. A positive ion can only replace another positive ion in a compoundA negative ion can only replace another negative ion in a compound

b) The element being replaced is less reactive than the element trying to replace ito Use Table J: Activity Series

- If the single element reactant is higher on Table J than the element it is trying to replace, a reaction WILL occur- If the single element reactant is lower on Table J than the element it is trying to replace, a reaction WILL NOT occur

Examples

For each of the following, determine whether or not a single replacement reaction will occur (TABLE J). If it does, complete the reaction and balance. If it does not occur, write “NO REACTION”.

1) ___Al + ___HCl ‡ ___ ________________ + ___ ________________

Al+3 + ___H+1Cl-1‡

The element Al is higher on Table J than the element H, so…..

Al+3 + H+1Cl-1‡ H2 + Al+3Cl3-1 ---leads to--- 2 Al + 6 HCl ‡ 3 H2 + 2 AlCl3

2) ___Ag + ___MgCl2‡ ___ ________________ + ___ ________________

Ag+1 + ___Mg+2Cl2-1‡

The element Ag is not higher on Table J than the element Mg, so…..

___Ag + ___MgCl2‡ NO REACTION

3) ___Cr + ___ Pb(NO3)2 ‡ ___ ________________ + ___ ________________

Cr+2 + ___Pb+2(NO3)2-1‡

The element Cr is higher on Table J than the element Pb, so…..

Cr+2 + Pb+2(NO3)2-1‡ Pb + Cr+2(NO3)2

-1 -leads to- Cr + Pb(NO3)2‡ Pb +Cr(NO3)2

4) ___Cl2 + ___ NaI ‡ ___ ________________ + ___ ________________

Cl2-1 + ___Na+1I-1‡

The element Cl is higher on Table J than the element I, so…..

Cl2-1 + Na+1I-1‡ I2 + Na+1Cl-1 ---leads to--- Cl2+ 2NaI ‡ I2 + 2 NaCl

5) Double Replacement ReactionsA reaction in which two compounds switch ions and form two new compound products.

General Formula: A+B- + C+D+- ‡ A+D- + C+B-

Rules to complete a double replacement reaction:a) Look up the charges of the elements in the compounds on the reactant sidesb) Positive ions will switch partners and form a bond with the other negative ion availablec) Negative ions will switch partners and form a bond with the other positive ion available.d) Use new subscripts if necessary to insure that each new formula has a net charge of zeroe) Determine the phases of the elements on Table F.

Soluble = aqueous, can dissolve in water, symbol is (aq)Insoluble = not aqueous, cannot dissolve in water

- a precipitate/solid will form, symbol is (s)

(s) (aq)(aq) (s)

This column contains This column containsnegative ions that when exceptions to the previous

This column contains This column contains in compounds will not column and if any of thesenegative ions that exceptions to the previous dissolve in water and are ions are also in the compoundalways form soluble column and if any of these insoluble (s) is now soluble (aq)compounds (aq) ions are also in the compound

the compound is now insoluble (s)

1) ___NaBr (aq) + ___AgNO3 (aq)‡ ___ ________________ + ___ ________________

___Na+1Br-1 (aq) + ___Ag+1NO3-1(aq)‡

Positive ion comes before negative ion in a compound, and opposites attract:

Na+1 will bond with NO3-1, and Ag+1 will bond with Br-1

Na+1Br-1 (aq) + Ag+1NO3-1(aq) ‡ NaNO3 ( ) + AgBr ( )

To determine phases of the two products, you must use reference table F:

Final Answer: 1 NaBr(aq) + 1 AgNO3(aq) ‡1 NaNO3 (aq) + 1 AgBr (s)

2) ___Ba(NO3)2 + ___Na2SO4 ‡ ___ ________________ + ___ ________________

___Ba+2NO3-1 (aq) + ___Na2

+1SO4-2(aq)‡

Positive ion comes before negative ion in a compound, and opposites attract:

Ba+2 will bond with SO4-2, and Na+1 will bond with NO3

-1

___Ba(NO3)2 (aq) + ___Na2SO4(aq)‡ BaSO4 ( ) + 2 NaNO3( )

To determine phases of the two products, you must use reference table F:

Final Answer: 1 Ba(NO3)2 (aq) + 1 Na2SO4(aq) ‡ 1 BaSO4 (s) + 2 NaNO3(aq)

3) ___AgNO3 (aq) + ___Na2CrO4 (aq) ‡ ___ ________________ + ___ ________________

___Ag+1NO3-1 (aq) + ___Na2

+1CrO4-2(aq)‡

Positive ion comes before negative ion in a compound, and opposites attract:

Ag+1 will bond with CrO4-2, and Na+1 will bond with NO3

-1

___AgNO3 (aq) + ___Na2CrO4(aq)‡ Ag2CrO4 ( ) + NaNO3( )

To determine phases of the two products, you must use reference table F:

Final Answer: 2 AgNO3 (aq) + 1 Na2CrO4(aq) ‡ 1 Ag2CrO4 (s) + 2 NaNO3(aq)

4) ___K2CO3 (aq) + ___CaSO4 (aq) ‡ ___ ________________ + ___ ________________

___K2+1CO3

-2 (aq) + ___Ca+2SO4-2(aq)‡

Positive ion comes before negative ion in a compound, and opposites attract:

K+1 will bond with SO4-2, and Ca+2 will bond with CO3

-2

___K2CO3(aq) + ___CaSO4(aq)‡ K2SO4 ( ) + CaCO3( )

To determine phases of the two products, you must use reference table F:

Final Answer: K2CO3(aq) + CaSO4(aq) ‡ K2SO4 (aq) + CaCO3(aq)

5) ___Mg(NO3)2 (aq) + ___LiOH ‡ ___ ________________ + ___ ________________

___Mg+2(NO3)2-1 (aq) + ___Li+1OH-1(aq)‡

Positive ion comes before negative ion in a compound, and opposites attract:

Mg+2 will bond with OH-1, and Li+1 will bond with NO3-1

___Mg(NO3)2 (aq) + ___LiOH ‡ ___Mg(OH)2 ( ) + ___LiNO3 ( )

To determine phases of the two products, you must use reference table F:

Final Answer: _1_Mg(NO3)2 (aq) + _2_LiOH ‡ _1_Mg(OH)2 (s) + _2_LiNO3 (aq)

E. Double Replacement Reactions and Net Ionic Equations

A net ionic equation shows only those particles involved in the formation of a solid (s) and is balanced with respect to mass and charge. You can predict the formation of a precipitate (solid) by using the general rules for solubility of ionic compounds with respect to Reference Table F.

When determining the net ionic equation, follow the steps used below in the example.

a) predict the products and balance the reaction; determine (s) and (aq) product

b) write the total balanced ionic equation, then cross out spectator ions (those ions that show up exactly the same on the reactant and product side)

c) write the balanced net ionic equation (what’s left over after you cross out the spectator ions)

- this is a reaction of the ions that create the solid precipitate

d) identify the spectator ions

Example 1:

Write the net ionic equation for the following reaction: Sr(NO3)2 (aq) + K2SO4 (aq)‡

a) Balance the Double Replacement Reaction: Sr(NO3)2(aq) + K2SO4 (aq)‡ 2 KNO3(aq) + SrSO4(s)

b) Write out the “Total Ionic Equation”:

Sr+2(aq) + 2 NO3

-1(aq) + 2 K+1

(aq) + SO4-2

(aq) ‡ 2 K+1(aq) + 2 NO3

-1(aq) + SrSO4(s)

c) Cross Out Spectator Ions – those ions that are the same on the reactant and product sides:

Sr+2(aq) + 2 NO3

-1(aq) + 2 K+1

(aq) + SO4-2

(aq) ‡ 2 K+1(aq) + 2 NO3

-1(aq) + SrSO4(s)

d) Net Ionic Equation – what’s left over: e) Spectator Ions – what was crossed out above:

Sr+2(aq) + SO4

-2(aq) ‡ SrSO4(s) 2 NO3

-1(aq) + 2 K+1

(aq)

Example 2:

Write the net ionic equation for the following reaction: KI (aq) + Pb(NO3)2 (aq)‡

a) Balance the Double Replacement Reaction: 2 KI (aq) + Pb(NO3)2 (aq) ‡ 2 KNO3(aq) + PbI2(s)

b) Write out the “Total Ionic Equation”:

2 K+1(aq) + 2 I-1

(aq) + Pb+2(aq) + NO3

-1(aq) ‡ 2 K+1

(aq) + 2 NO3-1

(aq) + PbI2(s)

c) Cross Out Spectator Ions – those ions that are the same on the reactant and product sides:

2 K+1(aq) + 2 I-1

(aq) + Pb+2(aq) + NO3

-1(aq) ‡ 2 K+1

(aq) + 2 NO3-1

(aq) + PbI2(s)

d) Net Ionic Equation – what’s left over: e) Spectator Ions – what was crossed out above:

Pb+2(aq) + 2I -1 (aq) ‡ PbI2(s) 2 NO3

-1(aq) + 2 K+1

(aq)

Notes: Stoichiometry

Stoichiometry is the study of amounts in chemical reactions. When doing stoichiometry problems, you must always begin with a balanced chemical reaction. You will then use the whole number coefficients as molar quantities for each compound, which will help to determine how much reactants are needed to produce a given amount of products.

Balanced Equations and Mole RatiosFrom a balanced equation, you can create mole ratios between two substances.

a) 2 H2 + 1 O2‡ 2 H2O b) 1 Zn + 2 HCl ‡ 1 H2 + 1 ZnCl2

Mole ratios: Mole ratios:2 moles H2 : 1 mole O2 1 mole Zn : 2 moles HCl2 moles H2 : 2 moles H2O 1 mole Zn : 1 mole H2

1 mole O2 : 2 moles H2 1 mole Zn : 1 mole ZnCl2

2 moles HCl : 1 mole H2

2 moles HCl : 1 mole ZnCl2

These mole ratios show the relationship between molar quantities of one compound to molar quantities of another compound

Stoichiometry Mole MapShows the relationship when converting between molar units AND between moles of one substance to moles of another substance.

use use22.4 L 22.4 L

use mole use

GFM ratio GFM

use use6.02 x 1023 6.02 x 1023

1) 1 step stoichiometry calculations (mole – mole calculations)

Liters of A Liters of B

Moles of A Moles of B

Atoms or Molecules

of A

Atoms or Molecules

of B

Grams of BGrams of A

This is the simplest type of stoichiometry calculation. These are mole-mole calculations, in which you are trying to determine how many moles of one substance there are related to another substance. (Use a one-step dimensional analysis setup to solve.) Using the mole ratio of the two substances being asked about, you will create a setup as below to solve these problems.

Example 1:Given the following balanced equation: 1 Zn + 2 HCl ‡ 1 H2 + 1 ZnCl2How many moles of H2 would be produced if there was 7.3 moles of HCl?

Given: 7.3 mole HCl x 1 mole H2 = 3.65 moles H2

1 2 moles HCl

mole ratio

Example 2:Given the following balanced equation: N2 + 3 H2‡ 2 NH3

How many moles of H2 are needed to produce 0.8 moles of NH3?

Given: 0.8 mole NH3 x 3 mole H2 = 1.2 moles H2

1 2 moles NH3

mole ratio

2) 2 – step stoichiometry calculationsIn a two-step calculation, you will be comparing moles of one substance to grams, liters or molecules of another substance in a balanced chemical equation. One step will be to convert the grams (or liters or atoms/molecule) to moles, and the other step will be to compare moles of one substance to moles of another substance using the mole ratios. This will require a two-stepdimensional analysis setup.

Example 1:Given the following balanced equation: 1 Zn + 2 HCl ‡ 1 H2 + 1 ZnCl2How many moles of H2 would be produced if there was 25.0 grams of Zn used?

Given: 25.0 g Zn x 1 mole Zn x 1 mole H2 = 0.306 moles H2

1 65.4 g Zn 1 mole Zn

GFM of Zn mole ratio

Example 2:

Given the following balanced equation: N2 + 3 H2‡ 2 NH3

How many liters of NH3 would be produced if there was 5.5 moles of N2?

Given: 5.5 Moles N2 x 2 moles NH3 x 22.4 L NH3 = 246.4 L NH3

1 1 mole N2 1 mole NH3

mole ratio from mole map

Example 3:Given the following balanced equation: 2 KClO3‡ 2 KCl + 3 O2

How many moles of KClO3 would be needed to produce 2.60 x 1023 molecules of O2?

Given: 2.60 x 1023 molecules O2 x 1 mole O2 x 2 moles KClO3 = 0.29 moles KClO3

1 6.02 x 1023 molecules 3 mole O2

from mole map mole ratio

3) 3 – step stoichiometry calculationsIn a three-step calculation, you will be comparing grams/atoms/molecules/liters of one substance to grams, liters, atoms or molecules of another substance in a balanced chemical equation. You will follow the mole map road and use a three-step dimensional analysis setup.

Example 1:Given the following balanced equation: 1 Zn + 2 HCl ‡ 1 H2 + 1 ZnCl2

How many grams of ZnCl2 would be produced if there was 37.0 grams of Zn used?

Given: 37.0 grams Zn x 1 mole Zn x 1 mole ZnCl2 x 136.4 g ZnCl2 = 77.2 g ZnCl2

1 65.4 g Zn 1 mole Zn 1 mole ZnCl2

Example 2:Given the following balanced equation: 1 N2 + 3 H2‡ 2 NH3

How many molecules of N2 would be needed to produce 125.0 L of NH3?

Given: 125.0 L NH3 x 1 mole NH3 x 1 mole N2 x 6.02 x 1023 molecules = 1.68 x 1024 molecules

1 22.4 L NH3 2 mole NH3 1 mole N2

Example 3:Given the following balanced equation: 2 KClO3‡ 2 KCl + 3 O2

How many liters of O2 would be produced if the reaction began with 20.0 grams of KClO3?

Given: 20.0 g KClO3 x 1 mole KClO3 x 3 mole O2 x 22.4 L O2 = 5.5 L O2

1 122.6 g KClO3 2 mole KClO3 1 mole O2

Limiting Reactants and Stoichiometry

The limiting reactant in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reactant, since the reaction cannot continue without it. If one or more other reactants are present in excess of the quantities required to react with the limiting reactant, they are described as excess reactants.The limiting reactant must be identified in order to calculate the percentage yield of a reaction, since the theoretical yield is defined as the amount of product obtained when the limiting reactant reacts completely. Given the balanced chemical equation, which describes the reaction, there are several equivalent ways to identify the limiting reactant and evaluate the excess quantities of other reactants.

Example #1: Limiting Reactant CalculationA 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. Which is the limiting reactant and how much excess reactant remains after the reaction has stopped?

First, we need to create a balanced equation for the reaction: 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

Next we can use stoichiometry to calculate how much product is produced by each reactant. NOTE: It does not matter which product is chosen, but the same product must be used for both reactants so that the amounts can be compared.

The reactant that produces the lesser amount of product in this case is the oxygen, which is thus the "limiting reactant." Next, to find the amount of excess reactant, we must calculate how much of the non-limiting reactant (ammonia) actually did react with the limiting reactant (oxygen).

We're not finished yet though. 1.70 g is the amount of ammonia that reacted, not what is left over. To find the amount of excess reactant remaining, subtract the amount that reacted from the amount in the original sample.

Example 2: Limiting Reactant Calculation90.0 g of FeCl3 reacts with 52.0 g of H2S. What is the limiting reactant? What is the mass of HCl produced? What mass of excess reactant remains after the reaction?

Percent Yield and Stoichiometry

The theoretical yield is the maximum amount of product you would expect from a reaction based on the amount of limiting reagent. In practice, however, chemists don’t always obtain the maximum yield for many reasons. Sincechemists know that the actual yield might be less than the theoretical yield, we report the actual yield using percentyield, which tells us what percentage of the theoretical yield we obtained. This ratio can be very valuable to otherpeople who might try your reaction. The percent yield is determined using the following equation:

Percent Yield = Actual yield x 100Theoretical yield

Since percent yield is a percentage, you would normally expect to have a percent yield between 0% and 100%. Ifyour percent yield is greater than 100%, that probably means you calculated or measured something incorrectly.

Problem #1: What is the percent yield of the following reaction if 60.0 grams of CaCO3 is heated to give 15.0grams of CaO?

CaCO3 --> CaO + CO2

Solution: Ideally, how many grams of CaO should be produced? First verify the equation is balanced; it is. Now determine the theoretical yield (in grams) of CaO, based on the amount of CaCO3 present.

60 grams CaCO3 x 1 mole CaCO3 x 1 mole CaO x 56.0 g CaO = 33.6 grams CaO1 100 grams CaCO3 1 mole CaCO3 1 mole CaO

So, ideally, 33.6 grams of CaO should have been produced in this reaction. This is the theoretical yield. However, the problem tells us that only 15 grams were produced. 15 grams is the actual yield. It is now a simple matter to find percent yield.

% Yield = 15.0 grams x 100 = 47.6 %33.6 grams

Problem #2: The following reaction is performed with 1.56 gram of BaCl2, and 1.82 grams of solid AgCl are produced.

BaCl2(aq) + 2 AgNO3 (aq) --> 2 AgCl(s) + Ba(NO3)2(aq)

What is the percent yield of the reaction?

1st: Determine the theoretical yield

1.56 grams BaCl2 x 1 mole BaCl2 x 2 mole AgCl x 143.3 g AgCl = 2.15 grams AgCl1 208.3 grams BaCl2 1 mole BaCl2 1 mole AgCl

So, ideally, 2.15 grams of AgCl should have been produced in this reaction. This is the theoretical yield. However, the problem tells us that only 1.82 grams were produced. 2.15grams is the actual yield. It is now a simple matter to find percent yield.

% Yield = 1.82 grams x 100 = 84.7 %2.15 grams