Solution Stoichiometry and Molarity

Solutions:

• Solutions are homogeneous mixtures composed of a solute and solvent
• Solute is the chemical present in lesser amount (whatever is dissolved)
• Solvent is the chemical present in the greater amount (whatever does the dissolving)
• Chemicals dissolved in water are aqueous
• NaCl(aq); H2O4(aq)

Molarity:

• Concentration can be expressed in many different ways
•  g/L, mL/L % by volume, % by mass, mol/L
• The most common one we use is mol/L which is also called Molarity
• mol/L = M
• [NaCl] = concentration of HCl

Example:

• 0.250mol of H2SO4 in 250 mL water

               0.250mol x 1/0.250L = 1mol/L

• 0.118mol of water in 2.50L

                0.118mol x 1/2.50L = 0.0472mol/L

Molarity Video:

- George Spencer

       

Limiting and Excess Reactants

What is a Limiting Reactant? 

In chemical reactions, usually one chemical gets used up before the other and this is known as the limiting reactant. The limiting reactant in a chemical reaction limits the amount of products that can be formed.  Additionally, the reaction will stop when all of the limiting reactant is consumed.

What is an Excess Reactant? 

The excess reactant in a chemical reaction remains when a reaction stops and when the limiting reactant is completely consumed.  The excess reactant remains because there is nothing with which it can react.

How to Find What is the Limiting Reactant and Excess Reactant in a Reaction?

Prior to solving a limiting reagent problem, you must: 

• write the balanced chemical reaction
• determine the stoichiometry of the reaction, or the relative proportion of components in the equation

Once you have a balanced equation is written and the stoichiometry is determined,  you will compare two  calculations to determine which reactant is the limiting reactant and which one is in excess and by what amount as outlined in the examples below. 

Example One: 

90.0 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?  

Balanced Chemical Reaction: 

• 2FeCl₃ + 3H₂S ----> 1Fe₂ S₃  +  6HCl

Example Two: 

3.50 g of Silver Nitrate reacts with 3.50 g of Copper (II) Chloride.  What is the limiting reactant?  What is the number of moles of excess reactant which remains after the reaction?    

Balanced Chemical Reaction: 

• 2AgNO₃ + 1CuCl₂ ----> 1AgCl₂ S₃  +  1Cu(NO₃)₂  

Limited Reactant:

• 3.50 g of AgNO₃   x  (1 mol/ 169.9 g)  x  (1/2)   x  (134.5 g/1 mol) = 1.39 grams of CuCl₂ (ENOUGH)
• 3.50 g of CuCl₂   x  (1 mol/ 134.5 g)  x  (2/1)   x  (169.5 g/1 mol) = 8.82 grams of AgNO₃ (NOT ENOUGH, AgNO₃ is the limiting reactant) 

Excess Reactant: 

CuCl₂ is the excess reactant so 

• 3.50  -  1.39   =   2.11  x  (1 mol/134.5g) = 0.0157 mol of CuCl₂
• 0.0157 mol of CuCl₂ = 1.57 x 10^-2 excess mol of CuCl₂

The following video should provide you with any additional or more thorough knowledge on the very important concept of limiting and excess reactants: 

 

- Simon Sierra

Percent Yield and Energy in Stoichiometry

Energy

• Enthalpy is the energy stored in chemical bonds
• Symbol of Enthalpy is H

                - units of joules (J) 

• Change in Enthalpy is ΔH 
• In exothermic reactions, enthalpy decreases
• In endothermic reactions, enthalpy increases

          

Calorimetry

• to experimentally determine the heat release we need to know 3 things

1. Tempurature change (ΔT)
2.  Mass (m)
3. Specific heat capacity (C)

• These are all related by the equation: ΔH = mCΔT

Example:

Calculate the heat required to warm up a cup of 400g of water. (C = 4.18J/g °C) from 20.0°C to 50.0°C.

 ΔH = mCΔT

        =(400g)(4.18J)(30°C)

        =50160 J

        =50.02 kJ

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?

%Yield = Actual/Theoretical

 Example:

The production of Urea: 2NH3 + CO2 → CO(NO2)2 + H2O

If 47.7g of Urea is produced, determine the theoretical yield of CO2.

What is the percent yield of CO2 if 12.0g is produced.

47.7g (mol/60g)(1/1)(44g/mol) = 35g

12.0/35 = 34%

-Benedict Suratos

Other Conversions: Volume to Heat

Volume at Standard Temperature and Pressure can be found using the conversion factor 22.4 L/mol. Heat can be individually included as a separate term in chemical reactions, known as enthalpy.  There are two distinctions with heat in chemical reactions: 

• Reactions that release heat (Exothermic) 
• Reactions that absorb heat (Endothermic) 

Both of these are seen in stoichiometry. 

Example One: 

If 5.0g of Potassium chlorate decompose according to the reaction below, what volume of oxygen gas (at STP) is produced? 

• 2KCLO3 ----> 2KCL + 3O2 
• 5.0 g x (1 mol/122.6 g) x (3/2) x (22.4L/1mol) = 1.4 L of O2 

Example Two: 

Find the amount of heat released when 5.0 mol of H2 are consumed according to the reaction:  

• N2 + 3H2 ---->  2NH3 + 46.2 kJ 
• 5.0 mol x (46.2 kJ/3) = 77 kJ

Find the amount of heat released when 250 g of Ammonia form according to the above reaction. 

• 250.0 mol x (1 mol/17.0g) x (46.2 kJ/2) = 3.4 x 10^2 kJ 

-Simon Sierra

Mass to Mass Problems

Mass to mass problems involve one additional conversion to what was done earlier. Because there is no direct link between the mass of one substance to the mass of another, as illustrated below, we must go through the moles and ratio to achieve an answer. 

 

The guide above would require us to do our equations as seen in the examples below. 

Example One:

Aluminum is produced by the decomposition of Bauxite. What mass of aluminum can be produced from 2.04 kg of Aluminum oxide?  

• 2Al2O3 ----> 4Al + 3O2 
• 2.04 Kg = 2040 g  

• 2040g x (1 mol/102g) x (4/2) x (27.0g/1mol) = 1080g ---> 1.08 x 10^3 g of Aluminum

Example Two:

Sulphuric Acid reacts with Potassium hydroxide to produce Potassium sulphate and water. What mass of Potassium sulphate can be produced from 294.6g of Potassium Hydroxide?  

• 2NaI + 1Pb(NO3)2----> 2NaNO3 + 1PbI2 

• 150g x (1 mol/85.0g) x (1/2) x (331.2g/1mol) = 292 g of Potassium Sulphate 

Here is a supplementary video: 

 

- Simon Sierra

Mole to Mass and Mass to Mole Problems

• Some Questions will give you an amount os moles and ask you to determine the mass
• Converting moles to mass only required one additional step

Example:

• When Silver reacts with 3.45 moles of Zinc phosphate what mass of Silver Phosphate would be produced?

                                               6Ag + ZN3(PO4)2 -> 2AG3(PO4) + 3Zn
                                              3.45mol x 2/1 x 418.7g/1mol = 2.89 x 10^3g

• Iron burns in air by reacting with Oxygen. If 5.6 g of Iron are completely burned, how many moles of Iron (III) Oxide are formed?

                                                                4Fe = 3O2 -> 2FE2O3

                                                        5.6g x 1mol/55.8g x 2/4 = 0.050mol

Moles to Mass:

Mass to Moles:

- George Spencer

Mole to Mole Problems

• Coefficients in balanced equations tell us the number of moles reacted or produced.
• They can also be used as conversion factors.
                                 3x + y -> 22
                                      2.0 mol
                                       1 : 2  <- ratio between y & z
  
  ratio between x & y -> 3:1
                                      6.0 mol
  
                             3x          +       y       ->    2z
                           [2.25 mol] [0.75 mol] I [1.5 mol]
                             3x          +      y        ->   2z
                           [1.2 mol]  x    [2/3]     =  [1.8 mol]
  
   
• Remember what you need over what you are given     

Example:

• Water decomposes into Hydrogen and Oxygen gas.
•  2H2O -> 2H2 = 1O2
• How many moles of Hydrogen gas are produced if 0.500 mol of water decompose?
• 0.500mol x 2/2 = 0.5000mol

Example:

• Sulphur reacts with Barium Oxide to produce Barium Sulphide and Oxygen gas.
• 1S8 + 8BaO -> 8Ba5 + 4O2
• How many moles of Sulphur are needed to completely react with 2.00 mol of Barium Oxide?
• 2.00mol x 1/8 = 0.250mol

Mole to Mole and Mass to Mass Conversions:

- George Spencer

Six Types of Reactions

• Understanding the 6 types of chemical reactions is the foundation of stoichiometry.
• Stoichiometry is a branch of chemistry that deals with the quantitative analysis of chemical reactions. 

The 6 types of reactions are:

1. Synthesis
2. Decomposition
3. Single Replacement
4. Double Replacement
5. Neutralization
6. Combustion

Synthesis A+B →AB

• Usually elements→compound
• Balanced synthesis reactions:

1. 2 Ag + S → Ag2S
2. 4 Fe + 3 O2 → 2 Fe2O3
3. N2 + 3 H2 → 2 NH3

Decomposition AB→ A+B

• Reverse of synthesis
• Assume the compounds decompose to elements during decomposition
• Balanced decomposition reactions:

1. 2H₂O  →    2H₂  +  O₂
2.  2NaCl →    2Na  +  Cl₂
3.  Mg₃N₂ → 3Mg + N₂

Single Replacement A+ BC → B + AC

• Balanced single replacement equations:

1. 2 AgNO3 + Cu → Cu(NO3)2 + 2 Ag 
2.  2 KI + Cl2 → 2 KCl + I2
3.  Cl₂ + 2NaBr → 2NaCl + Br₂

Double Replacement AB+CD → AD + BC

• Balanced double replacement equations:

1. AgNO3 + KCl → AgCl + KNO3
2.  HCl + NaOH → HOH + NaCl
3.  Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O

Neutralization 

• Reaction between an acid and a base
• Always produce water
• Balanced neutralization reactions: 

1. H₂SO₄ + 2 NaOH →  Na₂SO₄ + 2 H₂O
2.  HCl + NaOH → NaCl + HOH

Combustion

• Reaction when all substances in a compound are combined with oxygen which then produces carbon dioxide and water.
• Balanced combustion reactions:

1.  C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O 
2.  2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

-Benedict Suratos