Types of Reaction and Balancing Equation

Balancing Equations

The number of atoms the reactant side = the product side.

1.      Balance the atoms only occur in one molecule on each side of equation

2.      Don’t considering atoms separately, Balance whole groups whenever possible

3.       Don’t jump all over an equation, balancing a bit here and a bit there. Be systematic

4.      Balance atoms which occur in elemental form last. By elemental form we mean that the atoms are not combined with any atoms of a different kind

Li+ MgCl2 -> LiCl +Mg

Balance: 2Li+ MgCl2 -> 2LiCl +Mg

An equation isn’t properly balanced until we have actually balance both side

Ionic Compounds

Na SO4   Sodium Sulphate

K2Cl   Potassium Chloride

Fe2S3   Iron (III) Sulphide

Covalent Compound

1= mono, 7= hepta

2= di    8= octa

3= tri    9= nona

4= tetra  10= deca

5= penta 11= hendeca

6= hexa  12= dodeca

Acids

HCl= Hydro Chloric acid

H2SO4= Sulphurous acid

Types of Reactions

1.      Synthesis

It’s a reaction that combines 2 or more reactants to from one product

X+Y -> Z

Na+Cl-> NaCl

2.      Decomposition

It’s a reaction that breaks down one reactant in to 2 or more products

XY-> X+Y

NaCl-> Na+Cl

3.      Single replacement

It’s one where an element replaces an ion in a ionic compound. Metal replace positive ions (cations) and non metal replace negative ions (anions)

X+YZ-> XZ+Y (X= metal)

X+YZ-> ZY+Z (X= non mental)

2Al+ 3CuCl2-> 2AlCl3+ 3Cu

Predicting Single replacement reactions

Some metals are more reactive than other metals. So the metal has to be higher than the replace one to have reactions. Apply to both metals and non metal

Fe+ ZnCl2-> NR (because Fe isn’t more reactive than Zn)

3Mg+ 2AlCl3-> 3MgCl2+ 2Al (Yes because Mg is more reactive than Al)

4. Double replacement

A double replacement is a reaction between 2 ionic compounds usually in solution

WX+YZ-> WZ+XY

K2CO3+BaCl2-> 2KCl+ BaCO3

If the reactants change start during the reaction, this is a reaction occurring (usually precipitate forming)

If there’s no change of state, then there’s no reaction

Use “Table of Solubilities” to determine the states –(aq) or (s)

5. Combustion

It’s a reaction where burning in air is involved. The reactants are the chemical to be burned and the oxygen that it reacts with. The oxygen usually ends up combine more than one type of atoms as products.

AB+O2-> AO+BO

C4H8+6O2-> 4CO2+H2O

6. Neutralization

It’s a special double replacement reaction where acid react with bases to produce water

The acids have an H (+) as the cation and the bases have OH as the anion (-). Both should be (aq)

HA+BOH-> H2O+BA

        2HBr+Sr (OH) 2-> SrBr2+ 2H2O

Lab 4C: Formula of a hydrate

Objectives:
1.to determine the percentage of water in an unknown hydrate
2. to determine the moles of water present in each mole of this unknown hydrate, when given the molar mass of the anhydrous salt
3. to write the empirical formula of the hydrate.
Table1
Mass of empty crucible: 10.14g
Mass of crucible lid and hydrate: 11.20g
Mass of hydrate: 1.06g
Mass of crucible, lid and an hydrous salt(1st): 10.83g
Mass of crucible, lid and an hydrous salt(2nd): 10.81g
Mass of anhydrous salt: 0.67g
Mass of water given off: 0.39g
Mass of one mole of anhydrous salt from instructor: 159.6g/mol

1. 0.39g/1.06*100%= 36.8
2. 0.67g/159.6g/mol= 0.0042mol
3.0.39g/18g/mol=0.022mol
4. moles of water/ moles of CuSO4= 0.022mol/0.0042mol= 5
5.CuSO4 5H2O

Molar Volume at STP

Molar Volume of a Gas at STP

As you may know, gases change volume (expand and contract) with changes in pressure and temperature.

To compare volume of gases we use a standard condition called STP(Standard Temperature and Pressure)

STP = 1 atmosphere of pressure and a temperature of 0C (or) 273.15K

If at STP, 1 mole of gas occupies 22.4L, therefore we can create these conversion factors:
22.4L of gas
----------------
1 mole of gas

(or)

1 mole of gas
-----------------
22.4L of gas

Examples:
1. What volume will 3.20 moles of 02 gas occupy at STP?

3.20moles x 22.4L
1 mole

3.20moles x 22.4L “moles cancel, left with litres”
1 mole

= 71.7L occupied by 3.20moles 02 gas

2. What volume will 0.500 moles of NH3 gas occupy at STP?

0.500moles x 22.4L
1 mole

0.500moles x 22.4L “moles cancel, left with litres”
1 mole

= 11.2L occupied by 0.0500moles NH3 gas

3. What volume will 60.00g of CO2 gas occupy at STP?

28.0g x 1mole x 22.4L
28g 1mol


44.0g x 1mole x 22.4L = 30.55L occupied by 60g CO2
44g 1mole
“first convert to moles, then use moles to calculate litres occupied”
= 3630L occupied by 9.758 x 1025 atoms of N2O5

Diluting Solutions

Chemicals are shipped around the world is their most concentrated forms(solids, concentrated acids, etc)


For example.You have a 5.0 mg/L phosphate standard that exceeds the upper test limit for the low-range phosphate analytical procedure. For it to be considered adequate for a QA/QC known-concentration sample you’ll need to dilute it to a 0.2 mg/L concentration. Can you easily and accurately make this dilution?

Or your 6 N sodium hydroxide solution is too strong to properly adjust the pH of BOD samples. Instead, a 1 N solution would give greater control. How much of the 6 N solution would you need to dilute to get a 1 N solution?

There are also many other instances in a wastewater laboratory where the dilution of an acid, a base, or a laboratory standard is required so that when the resulting solution is used in an analytical procedure, it will help you attain the most accurate result.

The formula for diluting these types of solutions is simple:

(volumeA)(concentrationA) = (volumeB)(concentrationB)

Here are two examples using this formula:

#1) You want to make 250 mL of a 0.20 mg/L phosphate solution from a stock solution of 5.0 mg/L. How much of the 5.0 mg/L phosphate stock solution will you need to dilute to 250 mL so that the resulting concentration is 0.20 mg/L phosphate?

Put the information into the formula: (250 mL)(0.20 mg/L) = (X mL needed)(5.0 mg/L)
Then solve for X: (X mL needed) = (250 mL)(0.20 mg/L)/(5.0 mg/L) = 10 mL

Take 10 mL of the 5.0 mg/L phosphate solution and dilute it to 250 mL with distilled water (10 mL of solution with 240 mL water). The resulting solution will be 250 mL with a concentration of 0.20 mg/L phosphate.

Molarity

Molarity is the concentration of an element or compound in water. Molarity is known as M or mol/L.
The higher the molarity the more concentrated an element or compound is.

Molarity is easily calculated by finding how many moles are present and then dividing by the volume ( in litres). Simply if there are 2.5 moles present of a compound and the volume is 10 litres the molarity is 2.5/10 or .25 M.