Other Conversions:

·         Volume @ STP can be found using the conversion factor 22.4L/1mol

Ex.  If 6.0g of KCLO3 decompose according to the reaction below, what volume of O2 (@ STP) is produced?

1.      Balanced Eq’n: 2KClO3 à 2KCL + 3O2

2.      6.0g x 1mol/122.6g x 3/2 x 22.4L/1mol

= 1.6L

·         Heat can be included as a separate term in chemical reactions (enthalpy)…

·         Things to keep in mind when solving these eq’ns are that rxns that release heat are exothermic and rxns that absorb heat are endothermic. 

Ex.  Find the amount of heat released when 2.0mol of H2 are consumed accordingly to the reaction…

1.      Balanced Eq’n:  N2 +3H2 à 2NH3 + 46.2 kJ

2.      2.0mol x 46.2/3

=77kJ

Ex.  To find the amount of heat released when 675g of ammonia for according to the reaction below…

1.      Balanced Eq’n:  N2 + 3H2 à 2NH3 + 46.2kJ

2.      675g x 1mol/17g x 46.2/2

= 46.2kJ

Mole to Mass & Mass to Mole Conversions

Moles to Mass conversions…

·         Give you an amount of Moles and ask you to determine the mass.

·         Converting Moles – Mass only requires one additional step. 

Ex.  When silver reacts with 3.45 moles of zinc phosphate what mass of silver phosphate would be produced?

1.      Balanced Eq’n:  6Ag + Zn3(PO4)2 à 3Zn + 2Ag3PO4

2.      3.45mol x 2/1 x 418.7g/1mol

= 2889.03 ~ 2.89 x 10^3g

Ex.  You are given 4.00g of methane and are expected to burn it in order to initiate a chemical reaction.  How many moles of water vapour would be produced? 

1.      Balanced Eq’n:  CH4 + 2O2 à 2H2O + CO2

2.      4.00g x 1mol/16g x 2/1

= 0.500 mol

Mass-to-Mass conversions…

·         Involve one additional step to the “Moles to Mass” conversions

Ex.  When aluminum reacts with Iron (III) oxide to produce solid iron and aluminum oxide, what is the mass of aluminum oxide produced along with 19.55g of iron?

1.      Balanced Eq’n:  2Al + Fe2O3 à 2Fe + Al2O3

2.      19.55 x 1mol/55.8g x ½ x 102g/1mol

= 17.87g

·         To achieve the mass of aluminum oxide as shown above, we simply take its amu and multiply it by the final amount of moles. 

Ex.  Sodium iodide reacts with lead (II0 nitrate to produce sodium nitrate and lead (II) iodide.  What mass of lead (II) nitrate will be required to produce 150g of sodium nitrate?

1.      Balanced Eq’n:  2NaI + Pb(NO3)2 à NaNO3 + PbI2

2.      150g x 1mol/85g x ½ x 331.2/1mol

= 292g

Limiting Reactants

In Chemical reactions, usually one chemical gets used up first before the other. The chemical used up first in a chemical reaction is called the limiting reactant.

Once it is used up, the reaction stops! Limiting reactants determines the quantity of products formed.

To find the Limiting Reactant, assume one reactant is used up. Determine how much of this reaction is required!

Remember the very important equation for this:

Actual / Theoretical x 100% = % yield

Try some limiting reactant problems!

Energy and Percent Yield

• Enthalpy is the energy stored in chemical bonds
• The Symbol of Enthalpy is H, Units of Joules (J)
• Change in Enthalpy is ∆H
• In exothermic rnxs Enthalpy decreases
• in edothermic rnxs Enthalpy increases

Here are the definitions of exothermic and endothermic reactions, in case you forget:

 

EXOTHERMIC:

 

The term exothermic describes a process or reaction that releases energy from the system, usually in the form of heat, but also in the form of light, electricity or sound.

 

 

 

ENDOTHERMIC:

 

The word endothermic describes a process or reaction in which the system absorbs energy from the surroundings in the form of heat.

Calorimetry: To experimentally determine the heat released we need to know 3 things:

1. Temperature Change (∆T)
2. Mass (m)
3. Specific heat capacity (C)

These are related by the equation:

 

∆H = mC∆T

Example:

Calculate the heat required to warm a cup of 400 g of water (C= 41.8J/g°C) from 20.0°C to 50.0°C.

 

∆H = mC∆T

 

∆H =(400g)(4.18J/g°C)(50.0°C-20.0°C)

 

=50160 J

 

 

 

PERCENT YIELD

 

The theoretical yield of a reaction is the amount of products that SHOULD be formed. The actual amount depends on the experiment.

 

The percent yield is like a measure of success.

• how close is the actual amount to the predicted amount?

 

********THIS EQUATION IS VERY IMPORTANT*********

 

Actual / Theoretical x 100% = % yield

 

                      or

 

Actual amount of product / Expected amount of product x 100% = % yield

 

 

Example: A student makes a single displacement reaction that produces 2.755 grams of copper. He determines that 3.150 grams of copper should have been produced. Find the student's percentage yield.

 

 

                                      actual amount of product

percentage yield = ------------------------------------------- x 100

                                    expected amount of product

 

                                   2.755g

percentage yield = --------------- x 100

                                   3.150g

 

percentage yield = 87.46 %

 

 

 

 

 

 Try these percent yield questions!

 

 

Mass to Mass

Today we learned about Mass to Mass problems!

Example:

How many grams of O2 are produced from the decomposition of 3.0 g of Potassium Chlorate?


This question gives us grams, but we must change it to grams of another element. First we must write a balanced Chemical equation:

2KClO3 ----> 2K + Cl2 + 3O2


Than we take what we are given, which is 3.0 grams. Change 3.0 grams into moles, than find what you need over what you have, than with the moles you are given, find grams!


3.0 g x 1 mol/122.6 g = 0.024 mol x 3 O2 mol/ 2 KClO3 = 0.036 mol x 32.0 g/1mol
= 1.2 g

Let's try one more!

Example:

Determine the mass of lithium hydroxide produced when 0.38 g of lithium nitride reacts with water according to the following equation:

Li3N +3H2O ---> NH3+ 3LiOH


0.38 g x 1 mol/34.7 g = 0.011 mol x 23.95 g LiOH/1mol = 0.033 g x 23.9 g/ 1 mol
= 0.78 mol


Try some mass to mass problems, too!