Stoichiometric Calculations Stoichiometry Stoichiometric Calculations

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. I have 5 eggs. How many cookies can I make? Ratio of eggs to cookies 5 eggs 5 doz. 2 eggs = 12.5 dozen cookies

A. Proportional Relationships Stoichiometry relationships between substances in a chemical reaction based on the mole ratio Mole Ratio indicated by coefficients in a balanced equation 2 Mg + O2  2 MgO

What can we do with Stoichiometry? For the generic equation: RA +RB → P1 + P2 Given the… …one can find the… Amount of RA (or RB) Amount of RA (or RB) that is needed to react with it Amount of RA or RB Amount of P1 or P2 that will be produced Amount of P1 or P2 you need to produce Amount of RA &/or RB you must use

Given the equation… 2TiO2 + 4Cl2 + 3C → 2TiCl4 + CO2 + 2CO How many mol chlorine will react with 4.55 mol carbon? Given: 4.55 mole C Find: mole Cl2 6.07 mol Cl2 will react with 4.55 mol carbon

Given the same equation… 2TiO2 + 4Cl2 + 3C → 2TiCl4 + CO2 + 2CO What mass titanium (IV) oxide will react with 4.55 mol carbon? given = 4.55 mol C find = mass of TiO2 242.36 g TiO2 will react with 4.55 mol carbon

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

Standard Temperature & Pressure C. Molar Volume at STP 1 mol of a gas=22.4 L at STP Standard Temperature & Pressure 0°C and 1 atm

C. Molar Volume at STP LITERS OF GAS AT STP Molar Volume MASS IN GRAMS (22.4 L/mol) MASS IN GRAMS MOLES NUMBER OF PARTICLES Molar Mass (g/mol) 6.02  1023 particles/mol Molarity (mol/L) LITERS OF SOLUTION

D. Stoichiometry Problems How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas? 2KClO3  2KCl + 3O2 ? mol 9 mol 9 mol O2 2 mol KClO3 3 mol O2 = 6 mol KClO3

D. Stoichiometry Problems How many grams of KClO3 are req’d to produce 9.00 L of O2 at STP? 2KClO3  2KCl + 3O2 ? g 9.00 L 9.00 L O2 1 mol O2 22.4 L 2 mol KClO3 3 mol O2 122.55 g KClO3 1 mol KClO3 = 32.8 g KClO3

D. Stoichiometry Problems How many grams of silver will be formed from 12.0 g copper? Cu + 2AgNO3  2Ag + Cu(NO3)2 12.0 g ? g 12.0 g Cu 1 mol Cu 63.55 g Cu 2 mol Ag 1 mol Cu 107.87 g Ag 1 mol Ag = 40.7 g Ag

D. Stoichiometry Problems How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3? Cu + 2AgNO3  2Ag + Cu(NO3)2 ? g 1.5L 0.10M 1.5 L .10 mol AgNO3 1 L 1 mol Cu 2 mol AgNO3 63.55 g Cu 1 mol Cu = 4.8 g Cu

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 B3M2EE 30 B excess M and excess EE 10 B3M2EE 30 B and excess 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 W3P2SHF 50 S 50 W3P2SHF 50 S and excess of all other reactants

Solid aluminum reacts w/chlorine gas to yield solid aluminum chloride. 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 AlCl3

If 125 g chlorine react w/excess aluminum, how many g aluminum chloride are made? = 157 g AlCl3 If 125 g aluminum react w/125 g chlorine, how many g aluminum chloride are made? 157 g AlCl3 We’re out of Cl2.

limiting reactant (LR): the reactant that runs out first Amount of product is based on LR. 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, H, F Al / Cl2 / AlCl3 Cl2 Al

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

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.