Coach Hyde Physical Science Unit 5 Chemical Equations Coach Hyde Physical Science Unit 5

The Law of Conservation of Matter: states that for any system closed to all transfers of matter and energy (both of which have mass), the mass of the system must remain constant over time, as system mass cannot change quantity if it is not added or removed Soooo….. Matter cannot be created nor destroyed it can only change form (this is the basic principle of all conservation laws such as energy, matter, mass etc… Thank you Sir Antoine Lavoisier

(s): Solid stated of matter (l): Liquid State of matter Something to discuss before we get started (s): Solid stated of matter (l): Liquid State of matter (g): gaseous State of Matter (aq): compound is in aqueous solution with H2O

Balancing Chemical Equations

Reactants = The starting substances (things that react) Products = The final substances (produced) after the reaction

Exothermic Reaction: occurs when the temperature of a system increases due to the evolution of heat. This heat is released into the surroundings, A system of reactants that absorbs heat from the surroundings in an Endothermic Reaction. How do you think this can be noticed/determined?

If a chemical equation does not obey the law of conservation of mass the equation is said to be what? NOT BALANCED So Let’s look at the steps we need to take to BALANCE chemical equations Let’s work with the following equation: Fe + O2  Fe2O3

Step 1. Create a RAP table (what’s a RAP table ??) A table that shows us what atoms are present in this reaction, how many there are and are they reactants or products? For example: Fe + O2  Fe2O3 #R atom #P Fe 2 O 3

Rule 2. Go to the first atom that’s not balanced and balance it! Since Fe atoms are not balanced what do we need to do to balance it? Right! Multiply it by 2 (Only multiply) #R atom #P 1 Fe 2 2 O 3 2x

In step 2 we balanced the number of Fe atoms by multiplying the reactant side by 2. This now becomes the new coefficient in the chemical equation. Modify the equation to reflect the change #R atom #P 1 Fe 2 2 O 3 2Fe + O2  Fe2O3 2x Are all atoms balanced?

3. Move to the next unbalanced atom. What is it? #R atom #P 2 Fe 2 2 O 3 How can we balance the Oxygen? Multiply Reactants by 3 and Products by 2 Adjust the equation to reflect your changes #R atom #P 2 Fe 2 3x2 O 3x2 2Fe + 3O2  2Fe2O3 But notice that by changing Oxygen we also Changed Iron. We need to go back and fix this.

4. Write out the updated RAP table. How can we Balance the Iron? #R atom #P 2 Fe 4 6 O 6 Sure! Multiply the # of Reactant Fe atoms by 2 ! 2x Re-write the equation reflecting The new changes you’ve made. 4Fe + 3O2  2Fe2O3 Do we have a balanced Chemical Equation now? Yes we do!

Polyatomics #R atom #P 2 Ag 2 2 NO3 2 Mg 1 2 Cl 2 When an equation has Polyatomics in it, such as in this Balanced chemical equation 2AgNO3 + MgCl2  2AgCl + Mg(NO3)2 And the polyatomic appears on BOTH the reactant and product Side of the equation Count the polyatomic as an “ATOM” So the above reactant atoms would be: #R atom #P 2 Ag 2 2 NO3 2 Mg 1 2 Cl 2 If the same polyatomic does not Appear on both sides break the Polyatomic down into atoms!

Let's try solving some Problems

Is this equation balanced? NaOH + CaBr2  Ca(OH)2 + NaBr What atoms do we have in This equation? Count atoms & Start the RAP table #R atom #P NA 1 1 OH 2 Ca 1 2 Br 1 2) Do the #Reactant atoms = the # of Product atoms? 3) So pick the 1st unbalanced atom & begin balancing

We’ll start with balancing Hydroxide NaOH + CaBr2  Ca(OH)2 +NaBr How can we make both Hydroxides equal? #R atom #P 1 Na 1 1 OH 2 1 Ca 1 2 Br 1 Sure we’ll multiply #R OH by 2 Next step> rewrite the modified eqn. 2x 2NaOH + CaBr2  Ca(OH)2 + NaBr Hydroxide is now balanced so let’s move to the next Unbalanced atom, which is? …

What can we do to balance the Bromine? Sure! Multiply the #P Bromine by 2 Now adjust the table to reflect The changes and then rewrite the Eqn. #R atom #P 2 Na 1 2 OH 2 1 Ca 1 2 Br 1 x2 2NaOH + CaBr2  Ca(OH)2 + 2NaBr

Let’s update the RAP table with the new #’s Based on our updated equation. 2NaOH + CaBr2  Ca(OH)2 + 2NaBr #R atom #P 2 Na 2 2 OH 2 1 Ca 1 2 Br 2 Are we now balanced? Sure!

Ok Try Balancing this equation: C2H6 + O2  CO2 + H2O #R atom #P 2 C 1 6 H 2 2 O 3 Step 1. Total up the atoms Step 2. Balance the #P Carbon #R atom #P 2 C 2 6 H 2 2 O 5 & Re-write the equation C2H6 + O2  2CO2 + H2O Are we done?

Step 3. Carbons are balanced now but Hydrogen isn’t. So, balance Hydrogen atoms next #R atom #P 2 C 2 6 H 2 2 O 5 Multiply #P Hydrogen by 3 x3 Step 4. Re-write the eqn. & Retotal the number of atoms C2H6 + O2  2CO2 + 3H2O #R atom #P 2 C 2 6 H 6 2 O 7 Carbon and Hydrogen are now balanced but oxygen isn’t.

Step 5. To balance Oxygen multiply O by 3½ #R atom #P 2 C 2 6 H 6 2 O 7 Step 6. Re-write the eqn. & Retotal the number of atoms 3 ½ x C2H6 + 3½ O2  2CO2 + 3H2O It looks like we’re balanced. But, are we? No! We can’t have 3 ½ Oxygen molecules! Only whole Numbers are allowed. So what do we need to do to fix this?

Step 7. Let’s clean this up by Multiply everything by 2 C2H6 + 3½ O2  2CO2 + 3H2O x 2 2C2H6 + 7 O2  4CO2 + 6H2O #R atom #P 4 C 4 12 H 12 14 O 14 Step 8. Retotal #R and the #P atoms Are we balanced? YES!

Try this problem NH4OH + FeCl3  Fe(OH)3 + NH4Cl Start here. Recognize we Have polyatomics but they Appear on both sides of the Equation. #R atom #P 1 NH4 1 1 OH 3 1 Fe 1 3 Cl 1 OK … Now finish it up

Answer to previous problem 3NH4OH + FeCl3  Fe(OH)3 + 3NH4Cl

Types of Reactions You need to be able to identify each type. Synthesis reactions Decomposition reactions Single displacement reactions Double displacement reactions Combustion reactions You need to be able to identify each type.

1. Synthesis Example C + O2 O C +  O C General: A + B  AB

Ex. Synthesis Reaction

Practice Predict the products. Na(s) + Cl2(g)  Mg(s) + F2(g)  Al(s) + F2(g)  2 2 NaCl(s) MgF2(s) 2 3 2 AlF3(s) Now, balance them.

2. Decomposition Example: NaCl  Cl Na Cl + Na General: AB  A + B

Ex. Decomposition Reaction

Went from neutral (0) to (+2) Went from (+2) to Neutral (0) 3. Single Displacement Example: Zn + CuCl2 Zn was oxidized Went from neutral (0) to (+2)  Zn Cl Cu + Cl Zn Cu + Cu was reduced Went from (+2) to Neutral (0) General: AB + C  AC + B

Ex. Single Replacement Reaction

Single Replacement Reactions Write and balance the following single replacement reaction equation: Zn(s) + HCl(aq)  ZnCl2 + H2(g) 2 NaCl(s) + F2(g)  NaF(s) + Cl2(g) Al(s)+ Cu(NO3)2(aq) 2 2 2 3 3 Cu(s)+ Al(NO3)3(aq) 2

4. Double displacement General: AB + CD  AD + CB Example: MgO + CaS 

Double Replacement Reactions Think about it like “foil”ing in algebra, first and last ions go together + inside ions go together Example: AgNO3(aq) + NaCl(s)  AgCl(s) + NaNO3(aq) Another example: K2SO4(aq) + Ba(NO3)2(aq)  KNO3(aq) + BaSO4(s) 2

Practice Predict the products. HCl(aq) + AgNO3(aq)  CaCl2(aq) + Na3PO4(aq)  Pb(NO3)2(aq) + BaCl2(aq)  FeCl3(aq) + NaOH(aq)  H2SO4(aq) + NaOH(aq)  KOH(aq) + CuSO4(aq) 

5. Combustion Reactions Combustion reactions - a hydrocarbon reacts with oxygen gas. This is also called burning!!! In order to burn something you need the 3 things in the “fire triangle”: 1) Fuel (hydrocarbon) 2) Oxygen 3) Something to ignite the reaction (spark)

Combustion Reactions In general: CxHy + O2  CO2 + H2O Products are ALWAYS carbon dioxide and water. (although incomplete burning does cause some by-products like carbon monoxide) Combustion is used to heat homes and run automobiles (octane, as in gasoline, is C8H18)

Combustion Example C5H12 + O2  CO2 + H2O Write the products and balance the following combustion reaction: C10H22 + O2  8 5 6

Mixed Practice State the type & predict the products. BaCl2 + H2SO4  C6H12 + O2  Zn + CuSO4  Cs + Br2  FeCO3 