Section 6.5—Stoichiometry How can we determine in a lab the concentration of electrolytes?

What do those coefficients really mean? For every 2 moles of H2… and 2 moles of H2O are produced 2 2 2 H2 + O2  2 H2O No coefficient = 1 1 mole of O2 is need to react…

What is stoichiometry? Stoichiometry – Using the mole ratio from the balanced equation and information about one compound in the reaction to determine information about another compound in the equation.

Stoichiometry with Moles Example: If 4.2 mole of H2 reacts completely with O2, how many moles of O2 are needed? 2 H2 + O2  2 H2O

Stoichiometry with Moles Example: If 4.2 mole of H2 reacts completely with O2, how many moles of O2 are needed? 2 H2 + O2  2 H2O From balanced equation: 2 mole H2  1 mole O2 4.2 mole H2 1 mole O2 = ________ mole O2 2.1 2 mole H2

But we can’t measure moles in lab! We can’t go to the lab and count or measure moles…so we need a way to work in measurable units, such as grams and liters! Molecular mass gives the grams = 1 mole of a compound!

Stoichiometry with Moles & Mass Example: How many grams of AgCl will be precipitated if 0.45 mole AgNO3 is reacted as follows: 2 AgNO3 + CaCl2  2 AgCl + Ca(NO3)2

Stoichiometry with Moles & Mass Example: How many grams of AgCl will be precipitated if 0.45 mole AgNO3 is reacted as follows: 2 AgNO3 + CaCl2  2 AgCl + Ca(NO3)2 From balanced equation: 2 mole AgNO3  2 mole AgCl Molar Mass of AgCl: 1 mole AgCl = 143.35 g 0.45 mole AgNO3 2 mole AgCl 143.35 g AgCl = ________ g AgCl 65 2 mole AgNO3 1 mole AgCl

Stoichiometry with Mass Example: How many grams Ba(OH)2 are precipitated from 14.5 g of NaOH in the following reaction: 2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl

Stoichiometry with Mass Example: How many grams Ba(OH)2 are precipitated from 14.5 g of NaOH in the following reaction: 2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl Molar Mass of NaOH: 1 mole NaCl = 40.00 g From balanced equation: 2 mole NaOH  1 mole Ba(OH)2 Molar Mass of Ba(OH)2: 1 mole Ba(OH)2 = 171.35 g 14.5 g NaOH 1 mole NaOH 1 mole Ba(OH)2 171.35 g Ba(OH)2 40.00 g NaOH 2 mole NaOH 1 mole Ba(OH)2 = ________ g Ba(OH)2 31.1

But what about for solutions? Molarity gives the number of moles of the solute that are in 1 L of a solution

Stoichiometry with Solutions Example: If you need 15.7 g Ba(OH)2 to precipitate, how many liters of 2.5 M NaOH solution is needed? 2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl

Stoichiometry with Solutions Example: If you need 15.7 g Ba(OH)2 to precipitate, how many liters of 2.5 M NaOH solution is needed? 2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl Concentration of NaOH: 2.5 mole NaOH = 1 L From balanced equation: 2 mole NaOH  1 mole Ba(OH)2 Molar Mass of Ba(OH)2: 1 mole Ba(OH)2 = 171.35 g 15.7 g Ba(OH)2 1 mole Ba(OH)2 2 mole NaOH 1 L NaOH 171.35 g Ba(OH)2 1 mole Ba(OH)2 2.5 mole NaOH = ________ L NaOH 0.0733

What about gases? Standard Temperature and Pressure (STP) – 1 atm (or 101.3 kPa) and 273 K (0°C) Molar Volume of a Gas – at STP, 1 mole of any gas = 22.4 liters

Stoichiometry with Gases Example: If you need react 1.5 g of zinc completely, what volume of gas will be produced at STP? 2 HCl (aq) + Zn (s)  ZnCl2 (aq) + H2 (g)

Stoichiometry with Gases Example: If you need react 1.5 g of zinc completely, what volume of gas will be produced at STP? 2 HCl (aq) + Zn (s)  ZnCl2 (aq) + H2 (g) Molar volume of a gas: 1 mole H2 = 22.4 L From balanced equation: 1 mole Zn  1 mole H2 Molar Mass of Zn: 1 mole Zn = 65.39 g 1.5 g Zn 1 mole Zn 1 mole H2 22.4 L H2 65.39 g Zn 1 mole Zn 1 mole H2 = ________ L H2 0.51

Keeping all these equalities straight! TO GO BETWEEN USE THE EQUALITY Grams & moles Molecular Mass in grams = 1 mole moles & liters of a solution Molarity in moles = 1 L Moles & liters of a gas at STP 1 mole = 22.4 L at STP 2 different chemicals in a reaction Coefficient ratio from balanced equation

Titrations—Using Stoichiometry Titration – Addition of a known volume of a known concentration solution to a known volume of unknown concentration solution to determine the concentration.

End Point End Point (or Stoichiometric Point) – When there is no reactant left over—they have all be reacted and the solution contains only products The end point must be reached in order to use stoichiometry to calculate the unknown solution concentration Indicators – Paper or liquid that change color based on pH level. If the pH of the products is known, the indicator can be chosen to indicate the end point

Gravemetrics—Using Stoichiometry Gravemetric Analysis – Using a reaction to precipitate out an insoluble compound. The solid is dried and massed. Stoichiometry can then be used to determine the original substance’s concentration from the mass of the precipitate

If you are making 0.57 moles H2O, how many moles of O2 are needed? Let’s Practice #1 Example: If you are making 0.57 moles H2O, how many moles of O2 are needed? 2 H2 + O2  2 H2O

If you are making 0.57 moles H2O, how many moles of O2 are needed? Let’s Practice #1 Example: If you are making 0.57 moles H2O, how many moles of O2 are needed? 2 H2 + O2  2 H2O From balanced equation: 2 mole H2O  1 mole O2 0.57 mole H2O 1 mole O2 = ________ mole O2 0.29 2 mole H2O

2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl Let’s Practice #2 Example: If you need to precipitate 10.7 g of Ba(OH)2, how many grams NaOH are needed? 2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl

2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl Let’s Practice #2 Example: If you need to precipitate 10.7 g of Ba(OH)2, how many grams NaOH are needed? 2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl Molar Mass of Ba(OH)2: 1 mole Ba(OH)2 = 171.35 g From balanced equation: 2 mole NaOH  1 mole Ba(OH)2 Molar Mass of NaOH: 1 mole NaCl = 40.00 g 10.7 g Ba(OH)2 1 mole Ba(OH)2 2 mole NaOH 40.00 g NaOH 171.35 g Ba(OH)2 1 mole Ba(OH)2 1 mole NaOH = ________ g NaOH 5.00

Let’s Practice #3 Example: How many moles AgNO3 are needed to react with 10.7 g CaCl2? 2 AgNO3 + CaCl2  2 AgCl + 2 Ca(NO3)2

Let’s Practice #3 Example: How many moles AgNO3 are needed to react with 10.7 g CaCl2? 2 AgNO3 + CaCl2  2 AgCl + 2 Ca(NO3)2 From balanced equation: 2 mole AgNO3  1 mole CaCl2 Molar Mass of CaCl2: 1 mole CaCl2 = 110.98 g 10.7 g CaCl2 1 mole CaCl2 2 mole AgNO3 = ______ mole AgNO3 0.193 110.98 g CaCl2 1 mole CaCl2

2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl Let’s Practice #4 Example: How many liters of 0.10 M NaOH is needed to react with 0.125 L of 0.25 M BaCl2? 2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl

2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl Let’s Practice #4 Example: How many liters of 0.10 M NaOH is needed to react with 0.125 L of 0.25 M BaCl2? 2 NaOH + BaCl2  Ba(OH)2 + 2 NaCl Concentration of NaOH: 0.10 mole NaOH = 1 L From balanced equation: 2 mole NaOH  1 mole BaCl2 Concentration of BaCl2: 0.25 mole BaCl2 = 1 L BaCl2 0.125 L BaCl2 0.25 mole BaCl2 2 mole NaOH 1 L NaOH 1 L BaCl2 1 mole BaCl2 0.10 mole NaOH = ________ L NaOH 0.625

2 HCl (aq) + Zn (s)  ZnCl2 (aq) + H2 (g) Let’s Practice #5 Example: If you produce 15.4 L of H2 at STP, how many grams of ZnCl2 is also produced? 2 HCl (aq) + Zn (s)  ZnCl2 (aq) + H2 (g)

2 HCl (aq) + Zn (s)  ZnCl2 (aq) + H2 (g) Let’s Practice #5 Example: If you produce 15.4 L of H2 at STP, how many grams of ZnCl2 is also produced? 2 HCl (aq) + Zn (s)  ZnCl2 (aq) + H2 (g) Molar Mass of ZnCl2: 1 mole ZnCl2 = 136.29 g From balanced equation: 1 mole ZnCl2  1 mole H2 Molar volume of a gas: 1 mole H2 = 22.4 L 15.4 L H2 1 mole H2 1 mole ZnCl2 136.29 g ZnCl2 22.4 L H2 1 mole H2 1 mole ZnCl2 = ________ g ZnCl2 93.7