Stoichiometry Stoichiometric Calculations A Proportional Relationships 2 14

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Stoichiometry Stoichiometric Calculations

Stoichiometry Stoichiometric Calculations

A. Proportional Relationships 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt

A. Proportional Relationships 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. b. I have 5 eggs. How many cookies can I make? 5 eggs 5 doz. 2 eggs Ratio of eggs to cookies = 12. 5 dozen cookies

A. Proportional Relationships b Stoichiometry • relationships between substances in a chemical reaction •

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 Mg. O

What can we do with Stoichiometry? b For the generic equation: RA +RB →

What can we do with Stoichiometry? b For the generic equation: RA +RB → P 1 + P 2 Given the… …one can find the… Amount of RA (or RB) that is needed to react with it Amount of P 1 or P 2 that will be produced Amount of RA or RB Amount of P 1 or P 2 you need to produce Amount of RA &/or RB you must use

Given the equation… 2 Ti. O 2 + 4 Cl 2 + 3 C

Given the equation… 2 Ti. O 2 + 4 Cl 2 + 3 C → 2 Ti. Cl 4 + CO 2 + 2 CO 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 will react with 4. 55 mol carbon

Given the same equation… 2 Ti. O 2 + 4 Cl 2 + 3

Given the same equation… 2 Ti. O 2 + 4 Cl 2 + 3 C → 2 Ti. Cl 4 + CO 2 + 2 CO b What mass titanium (IV) oxide will react with 4. 55 mol carbon? b given = 4. 55 mol C b find = mass of Ti. O 2 b 242. 36 carbon g Ti. O 2 will react with 4. 55 mol

B. Stoichiometry Steps 1. Write a balanced equation. 2. Identify the given and find.

B. Stoichiometry Steps 1. Write a balanced equation. 2. Identify the given and find. 3. Line up conversion factors. • Mole ratio - moles • Molar mass - moles grams • Molarity - moles liters soln • Molar volume - moles liters gas Core step in all stoichiometry problems!! 4. Check answer.

C. Molar Volume at STP 1 mol of a gas=22. 4 L at STP

C. Molar Volume at STP 1 mol of a gas=22. 4 L at STP Standard Temperature & 0°C and 1 atm Pressure

C. Molar Volume at STP LITERS OF GAS AT STP Molar Volume (22. 4

C. Molar Volume at STP LITERS OF GAS AT STP Molar Volume (22. 4 L/mol) MASS IN GRAMS Molar Mass (g/mol) 6. 02 MOLES 1023 particles/mol Molarity (mol/L) LITERS OF SOLUTION NUMBER OF PARTICLES

D. Stoichiometry Problems b How many moles of KCl. O 3 must decompose in

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

D. Stoichiometry Problems b How many grams of KCl. O 3 are req’d to

D. Stoichiometry Problems b How many grams of KCl. O 3 are req’d to produce 9. 00 L of O 2 at STP? 2 KCl. O 3 2 KCl + 3 O 2 ? g 9. 00 L O 2 1 mol 2 mol 122. 55 O 2 KCl. O 3 g KCl. O 3 = 32. 8 g 22. 4 L 3 mol 1 mol KCl. O 3 O 2 KCl. O 3

D. Stoichiometry Problems b How many grams of silver will be formed from 12.

D. Stoichiometry Problems b How many grams of silver will be formed from 12. 0 g copper? Cu + 2 Ag. NO 3 2 Ag + Cu(NO 3)2 12. 0 g ? g 12. 0 1 mol 2 mol 107. 87 g Cu Cu Ag g Ag = 40. 7 g 63. 55 1 mol Ag g Cu Cu Ag

D. Stoichiometry Problems b How many grams of Cu are required to react with

D. Stoichiometry Problems b How many grams of Cu are required to react with 1. 5 L of 0. 10 M Ag. NO 3? Cu + 2 Ag. NO 3 2 Ag + Cu(NO 3)2 ? g 1. 5 L 0. 10 M 1. 5. 10 mol 1 mol 63. 55 L Ag. NO 3 Cu g Cu 1 L = 4. 8 g 2 mol 1 mol Cu Ag. NO 3 Cu

The Limiting Reactant A balanced equation for making a Big Mac® might be: 3

The Limiting Reactant A balanced equation for making a Big Mac® might be: 3 B + 2 M + EE B 3 M 2 EE 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

A balanced equation for making a tricycle might be: 3 W + 2 P + S + H + F W 3 P 2 SHF 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. Al(s) + 3 Cl

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

If 125 g chlorine react w/excess aluminum, how many g aluminum chloride are made?

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

blimiting reactant (LR): the reactant that runs out first • Amount of product is

blimiting 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 Macs buns meat tricycles pedals W, S,

Example Limiting Reactant Excess Reactant( s) Big Macs buns meat tricycles pedals W, S, H, F Al / Cl 2 / Al. Cl 3 Cl 2 Al

How to Find the Limiting Reactant b. For the generic reaction RA + RB

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

How to find LR steps… 1. Calc. # of mol of RA and RB

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