Stoichiometric Calculations Stoichiometry. A. Proportional Relationships b I have 5 eggs. How many cookies can I make? 3/4 c. brown sugar 1 tsp vanilla.

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Presentation transcript:

Stoichiometric Calculations Stoichiometry

A. Proportional Relationships b I have 5 eggs. How many cookies can I make? 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 5 eggs5 doz. 2 eggs = 12.5 dozen cookies Ratio of eggs to cookies

A. Proportional Relationships b Stoichiometry relationships between substances in a chemical reaction based on the mole ratio b Mole Ratio indicated by coefficients in a balanced equation 2 Mg + O 2  2 MgO

What can we do with Stoichiometry? b For the generic equation: R A +R B → P 1 + P 2 Given the……one can find the… Amount of one reactantAmount of other reactant that is needed to react with it Amount of reactantAmount of product that will be produced Amount of product you need to produce Amount of reactant you must use

B. Stoichiometry Steps 1. Write a balanced equation. 2. Identify the given and find. 3. Line up conversion factors. Mole ratio - moles  moles Convert amount of Substance A into moles (if needed) Use a Mole ratio (must be included in ALL stoichiometry problems) Convert moles of Substance B into desired unit 4. Check answer.

Given the equation… 2TiO 2 + 4Cl 2 + 3C → 2TiCl 4 + CO 2 + 2CO b How many mol chlorine will react with 4.55 mol carbon? b Given: 4.55 mole C b Find: mole Cl 2 b 6.07 mol Cl 2

Given the same equation… 2TiO 2 + 4Cl 2 + 3C → 2TiCl 4 + CO 2 + 2CO b What mass titanium (IV) oxide will react with 4.55 mol carbon? b given = 4.55 mol C b find = mass of TiO 2 b g TiO 2

D. Stoichiometry Problems b How many moles of KClO 3 must decompose in order to produce 9 moles of oxygen gas? 9 mol O 2 2 mol KClO 3 3 mol O 2 = 6 mol KClO 3 2KClO 3  2KCl + 3O 2 ? mol9 mol

b How many grams of KClO 3 are req’d to produce 9.00 L of O 2 at STP? 9.00 L O 2 1 mol O L O 2 = 32.8 g KClO 3 2 mol KClO 3 3 mol O g KClO 3 1 mol KClO 3 ? g9.00 L D. Stoichiometry Problems 2KClO 3  2KCl + 3O 2

D. Stoichiometry Problems b How many grams of silver will be formed from 12.0 g copper? 12.0 g Cu 1 mol Cu g Cu = 40.7 g Ag Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 2 mol Ag 1 mol Cu g Ag 1 mol Ag 12.0 g? g

63.55 g Cu 1 mol Cu D. Stoichiometry Problems b How many grams of Cu are required to react with 1.5 L of 0.10M AgNO 3 ? 1.5 L.10 mol AgNO 3 1 L = 4.8 g Cu Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 1 mol Cu 2 mol AgNO 3 ? g 1.5L 0.10M

The Limiting Reactant A balanced equation for making a Big Mac® might be: 3 B + 2 M + EE  B3M2EE With……and… …one can make… 30 M excess B and excess EE 15 B 3 M 2 EE 30 B excess M and excess EE 10 B 3 M 2 EE 30 M 30 B and excess EE 10 B 3 M 2 EE

A balanced equation for making a tricycle might be: 3 W + 2 P + S + H + F  W3P2SHF With……and… …one can make… 50 P excess of all other reactants 25 W 3 P 2 SHF 50 S excess of all other reactants 50 W 3 P 2 SHF 50 P 50 S and excess of all other reactants 25 W 3 P 2 SHF

Solid aluminum reacts w/chlorine gas to yield solid aluminum chloride. 2 Al(s) + 3 Cl2(g)  2 AlCl3(s) If 125 g aluminum react w/excess chlorine, how many g aluminum chloride are made? = 618 g AlCl 3

If 125 g chlorine react w/excess aluminum, how many g aluminum chloride are made? = 157 g AlCl 3 b If 125 g aluminum react w/125 g chlorine, how many g aluminum chloride are made? 157 g AlCl 3 b We’re out of Cl 2.

b limiting reactant (LR): the reactant that runs out first Amount of product is based on LR. b Any reactant you don’t run out of is an excess reactant (ER).

Example Limiting Reactant Excess Reactant(s) Big Macsbunsmeat tricyclespedalsW, S, H, F Al / Cl 2 / AlCl 3 Cl 2 Al

How to Find the Limiting Reactant b For the generic reaction R A + R B  P b Assume that the amounts of R A and R B are given. b Should you use R A or R B in your calculations?

How to find LR steps… 1. Calc. # of mol of R A and R B you have. 2. Divide by the respective coefficients in balanced equation. 3. Reactant having the smaller result is the LR.