Calculations with Limiting Reagents

• Limiting reactant – reactant (reagent) that is used up first and thus limits how much product can be made

• The other reactant is said to be in excess

2 eggs + 1 bag chips 50 cookies

1. Stoichiometric quantities = combine 2 eggs and 1 bag of chips, make 50 cookies

2. Have 4 eggs and 1 bag of chips:• Limiting reagent: chip• Reagent in excess: eggs• Can make _____ cookies

• Answer: 50 cookies

N2 + 3 H2 2 NH3

5.00 grams of N2 and 2.50 grams of H2 are combined and allowed to react.

1. Which reactant is limiting and which is in excess?

– Limiting: N2 (results in 6.07 g NH3)

– Excess: H2 (would result in 14.0 g NH3if all reacted)

2. How many grams of NH3 will be made?

– 6.07 g NH3

Zn + 2 AgNO3 2 Ag + Zn(NO3)2

• A piece of zinc weighing 2.00 grams is placed in a solution containing 2.50 grams of silver nitrate. How many grams of silver (Ag) will form?

– First need to determine which reagent is limiting and which is in excess

– Limiting: AgNO3 1.59 g Ag

– Excess Zn

Percent Yield

• The amount of product collected after a reaction is often less than the maximum amount possible.

• Why might this occur?

Percent Yield

• Percent yield is a measure of what % of the maximum yield is actually made or collected.– Measure of efficiency of the reaction or

process

% Yield = actual yield x 100%

theoretical yield

Zn + 2 AgNO3 2 Ag + Zn(NO3)2

• A piece of zinc weighing 2.00 grams is placed in a solution containing 2.50 grams of silver nitrate. Upon completion of the reaction 1.43 grams of silver are collected. What is the percent yield for this reaction?

– Based on slide 4….• %yield= 1.43 g/1.59 g x 100% = 89.9%

4 NH3 + 5 O2 4 NO + 6 H2O

1. How many grams of NO are formed when 1.50 grams of NH3 reacts with 1.00 grams of O2?

• Which reactant is limiting and which is in excess?

• Limiting O2 make 0.750 g NO

• Excess: NH3 – would make 2.65 g NO if all could react

• How much of the reactant in excess will be left over after the reaction is complete?

• Leftover NH3 = 1.50 g - .425 g => 1.08 g

CH 4: Types of Reactions, Solution Stoichiometry

• Water as solvent

• Electrolytes

• Solution composition – Molarity

• Molarity calculations

• Reactions in solution

• Types of reactions - lab

Water as solvent

• Water can dissolve:– small polar molecules

• Water molecules are attracted to the polar molecules

– many ionic compounds • Water molecules hydrate the ions – this separates

the ions and allows them to move freely through the water

• Water is a polar molecule with areas of positive and negative charge

H H

O

+ +

-

Hydrogen bonding in water

Polar molecules can H bond to the water molecules (see board).

Ionic Bonding

Water Hydrates Ions

Electrolytes

• Electrolyte – substance that conducts electricity when dissolved in water– To conduct electricity there must be charged

particles free to move– This occurs when an ionic compund dissolves

in water.

Types of Electrolytes

• Strong electrolytes – substances that ionize completely in water

– Soluble salts– Strong acids

• HCl, HNO3, H2SO4

– Strong bases• Group IA hydroxides - NaOH, KOH

Types of Electrolytes

• Weak electrolytes – substances that ionize slightly in water

– Few of the dissolved particles form ions– Weak acids

• All acids but: HCl, HNO3, H2SO4

• HCN, HC2H3O2

– Weak bases• Often have N in their formula

• NH3

Types of Electrolytes

• Nonelectrolytes – substances that dissolve in water*, but do not form ions– Polar molecular compounds

• Alcohols• Sugars

* others define as any substance that does not conduct electricity

Concentration Unit - Molarity

• Molarity = moles solute

L solution

Molarity Calculations

1. Molarity of a given solution

2. How to make a specific volume of solution of specific M

3. Calculations using M

4. Stoichiometry in aqueous solution

Making Solutions• Describe how to make 1.00 L of a 2.00 M

NaOH solution.

• Describe how to make 500. mL of a 2.00M NaOH solution.

• Describe how to make 375 mL of a 0.150M Pb(NO3)2 solution.

Molarity of Solutions

• What is the molarity of a solution made by dissolving 17.00 g of NaOH in enough water to make 250. mL of solution?

• What is the molarity of a solution made by dissolving 5.00 g NaCl in enough water to make 75.0 mL of solution?

From Molarity to Moles

• Molarity links moles and volume.

• If you know 2 of these you can calculate the 3rd.

• We can use molarity as a conversion factor or take the algebraic approach.

M = mol

L

Molarity Calculations

• How many moles of HCl are in 125 mL of 3.00 M HCl?

• What volume of 1.50 M H2SO4 contains 0.325 moles of H2SO4 ?

Reaction Stoichiometry

• Mg + 2 HCl MgCl2 + H2

• How many mL of 6.0 M HCl are required to fully react 0.40 grams of Mg?

Reaction Stoichiometry

• NaOH + HCl NaCl + H2O

It takes 13.0 mL of 0.200 M NaOH to react 5.00 mL of HCl solution. What is the concentration of the HCl solution?

Reaction Stoichiometry

• H2SO4 + 2 NaOH products

• It takes 23.0 mL of 0.200 M NaOH to react 7.00 mL of H2SO4 solution. What is the concentration of the H2SO4 solution?