# III. Stoichiometry Stoy – kee – ahm –eh - tree

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III. Stoichiometry Stoy – kee – ahm –eh - tree
Chapter 12

Mole Ratios Mole-to-Mole Calculations Mole-to-Mass Calculations Particle Calculations Molar Volume Calculations Limiting Reactants Percent Yields

Things you should remember
From the Moles Unit: Identify particles as atoms, molecules (mc), and formula units (fun) 1 mole = 6.02 x 1023 atoms, molecules, or formula units 1 mole substance = mass (in grams) from the periodic table From the Naming & Formulas Unit: How to write a formula given a chemical name From the Chemical Reactions Unit: How to write a chemical equation given words Balancing equations

I. Stoichiometry stoikheion, meaning element metron, meaning measure
Thus Stoichiometry- measuring elements!

III. Stoichiometry Stoichiometry (sometimes called reaction stoichiometry to specify its use in analyzing a chemical reaction) is the calculation of quantitative (numbers) relationships between the reactants and products in a balanced chemical reaction.

A. Basis for Calculations
The basis for properly working stoichiometry problems is the Balanced Chemical Equation and the Mole Ratio. These are VITAL TO YOUR SURVIVAL IN STOICHIOMETRY!!!

1. Review of a Balanced Chemical Equation
You must always check to ensure you have a proper chemical equation. It is not correctly balanced you can not use it for any calculations!!!!

Ex: Properly Balance the following equation
2 __H2SO4(aq) + __NaHCO3(s)  __Na2SO4(aq) + __H2O(l) + __CO2(g) 2 2

2. Molar Ratio Once the equation is properly balanced, there are relationships between all of the compounds involved. These are called MOLE RATIOS. any two compounds can be written as a relationship in terms of moles

For the Reaction: H2SO4 + 2NaHCO3 Na2SO4 + 2H2O + 2CO2
What’s the relationship between NaHCO3 and Na2SO4? 2 mole NaHCO3 = 1 mol Na2SO4 (it’s just the coefficients!!) What’s the relationship between H2SO4 and CO2? 1 mol H2SO4 = 2 mol CO2

For the Reaction: H2SO4 + 2NaHCO3 Na2SO4 + 2H2O + 2CO2
1 mol H2SO4 = 2 mol NaHCO3 or 1 mc H2SO4 = 2 mc NaHCO3 1 mol/mc H2SO4 2 mol/mc NaHCO3 1 mole/mc H2SO4

For the Reaction: H2SO4 + 2NaHCO3 Na2SO4 + 2H2O + 2CO2
1 mol H2SO4 = 2 mol CO2 or 1 mc H2SO4 = 2 mc CO2 1 mol/mc H2SO4 2 mol/mc CO2 1 mole/mc H2SO4

Example: Iron reacts with oxygen to create iron(III) oxide.
After writing a balanced equation, write down 3 possible relationships. (hint: just look at the coefficients.) Skeletal equation: Fe + O2  Fe2O3 Balanced equation: 4Fe + 3O2  2Fe2O3 Relationships: 4 mol Fe = 3 mol O2 4 mol Fe = 2 mol Fe2O3 3 mol O2 = 2 mol Fe2O3

Practice 6 2 3 Al2S3 + H2O  Al(OH)3 + H2S 1 mol Al2S3 = 6 mol H2O
1. Write three mole ratios (relationships) from the reaction below: Al2S H2O  Al(OH) H2S 6 2 3 1 mol Al2S3 = 6 mol H2O 1 mol Al2S3 = 2 mol Al(OH)3 1 mol Al2S3 = 3 mol H2S You should have 3 6 mol H2O = 2 mol Al(OH)3 6 mol H2O = 3 mol H2S 2 mol Al(OH)3 = 3 mol H2S

Practice 1. Write three mole ratios (relationships) from the reaction below: Al2S H2O  Al(OH) H2S 6 2 3 1 mol Al2S3 = 6 mol H2O 1 mol Al2S3 = 2 mol Al(OH)3 1 mol Al2S3 = 3 mol H2S 6 mol H2O = 2 mol Al(OH)3 6 mol H2O = 3 mol H2S

Practice 2Al2O3  4Al + 3O2 2 mol Al2O3 = 4 mol Al
2. Aluminum is produced by decomposing aluminum oxide into aluminum and oxygen. a. Write a balanced equation. b. Write all the molar ratios that can be derived from this equation. 2Al2O3  4Al + 3O2 2 mol Al2O3 = 4 mol Al 2 mol Al2O3 = 3 mol O2 4 mol Al = 3 mol O2

NOTE: NOTE: Every time you do a stoichiometric calculation you MUST use a mole ratio. The mole ratio allows you to compare one compound in an equation with another. Don’t forget to balance your chemical equations!!!

B. Calculating Problems

B. Calculating Problems - Stoich it up!
Before any stoich problem you have to set it up. Consider this the pre-game warm-up. This should become second nature to you.

Pre-game Warm-up: 1. Write a balanced reaction.
2. Determine your given & want 3. Determine your relationships. If you see… 2 different substances, determine their mole ratio mass (g, mg, kg), calculate the molar mass of that substance atoms, molecules, or fun, 1 mol = 6.02 x 1023 of that type extras like mg or kg, you know what to do Now you are ready to solve- IT’S GAME TIME.

Game Time: put your GIVEN OVER 1
place your relationships where the units cancel out diagonally everything equal to each other goes above and below each other cancel out your units until you are left over with your wanted

Here is a flow chart that we will dissect this unit to do our problems

For most of the examples we will be using this equation:
Haber Process: N2 (g) H2 (g)  2NH3 (g) Haber Process: an industrial process for producing ammonia from nitrogen and hydrogen by combining them under high pressure in the present of an iron catalyst source: worldnet.princeton.edu

1. Moles to Moles

4.00 mol H2 2 mol NH3 = mol NH3 2.67 3 mol H2 1 Balanced Equation:
Ex 1: How many moles of ammonia are produced from 4.00 moles of hydrogen gas in the presence of excess nitrogen? Balanced Equation: N2 (g) H2 (g)  2NH3 (g) No grams, No molar mass!! Given : Want : Relationships: 4.00 mol H2 No fun, mc, No 6.02 x 1023!! ? mol NH3 2 mol NH3 = 3 mol H2 4.00 mol H2 2 mol NH3 = mol NH3 2.67 x 3 mol H2 1

7.8 mol NH3 1 mol N2 = mol N2 3.9 2 mol NH3 1 Balanced Equation:
Ex 2: How many moles of nitrogen were used if 7.8 moles of ammonia were made in excess hydrogen gas? Balanced Equation: N2 (g) H2 (g)  2NH3 (g) No grams, No molar mass!! Given : Want : Relationships: 7.8 mol NH3 No fun, mc, No 6.02 x 1023!! ? mol N2 2 mol NH3 = 1 mol N2 7.8 mol NH3 1 mol N2 = mol N2 3.9 X 2 mol NH3 1

Ex 3: How many moles of hydrogen react with 13 moles of nitrogen?
Balanced Equation: N2 (g) H2 (g)  2NH3 (g) No grams, No molar mass!! Given : Want : Relationships: 13 mol N2 No fun, mc, No 6.02 x 1023!! ? mol H2 1 mol N2 = 3 mol H2 13 mol N2 3 mol H2 = moles H2 39 x 1 mol N2 1

2. Moles to Mass/ Mass to Moles

Steps to Success! 1. Write a balanced reaction.
2. Determine your given & want 3. Determine your relationships. If you see… 2 different substances, determine their mole ratio mass (g, mg, kg), calculate the molar mass of that substance atoms, molecules, or f.un, 1 mol = 6.02 x 1023 of that type extras like mg or kg, you know what to do

Ex 1: How many grams of ammonia are produced from 4
Ex 1: How many grams of ammonia are produced from 4.00 moles of hydrogen gas in excess nitrogen? Balanced Equation: N2 (g) H2 (g)  2NH3 (g) How do I know when to use mole ratios, molar mass or 6.02 x 1023? Given : Want : Relationships: 4.00 mol H2 ? g NH3 2 mol NH3 = 3 mol H2 1 mol NH3 = g NH3 Look at the “given” and “want” for clues No fun, mc, No 6.02 x 1023!!

4.00 mol H2 2 mol NH3 17.031g NH3 3 mol H2 1 mol NH3 1 = g NH3 45.4
Ex 1: How many grams of ammonia are produced from 4 moles of hydrogen gas in excess nitrogen? Balanced Equation: N2 (g) H2 (g)  2NH3 (g) Given : Want : Relationships: 4.00 mol H2 ? g NH3 2 mol NH3 = 3 mol H2 1 mol NH3 = g NH3 4.00 mol H2 2 mol NH3 17.031g NH3 x x 3 mol H2 1 mol NH3 1 = g NH3 45.4

Ex 2: How many moles of ammonia are produced from 4.00 grams of hydrogen gas in excess nitrogen? Balanced Equation: N2 (g) H2 (g)  2NH3 (g) Given : Want : Relationships: 4.00 g H2 ? mol NH3 2 mol NH3 = 3 mol H2 1 mol H2 = 2.016g H2 Remember your Steps to Success! No fun, mc, No 6.02 x 1023!!

4.00 g H2 1 mol H2 2 mol NH3 2.016g H2 3 mol H2 1 = mol NH3 1.32
Ex 2: How many moles of ammonia are produced from 4 grams of hydrogen gas in excess nitrogen? Balanced Equation: N2 (g) H2 (g)  2NH3 (g) Given : Want : Relationships: 4.00 g H2 ? mol NH3 2 mol NH3 = 3 mol H2 1 mol H2 = 2.016g H2 4.00 g H2 1 mol H2 2 mol NH3 x x 2.016g H2 3 mol H2 1 = mol NH3 1.32

Practice 1. How many grams of nitrogen will react with 3.40 moles of hydrogen to produce ammonia? 2. How many grams of hydrogen are required to make 54.0 moles of ammonia in excess nitrogen? 3. How many moles of nitrogen react completely with 3.70 moles of hydrogen? 31.7 g N2 163 g H2 1.23 mol N2

3. Mass to Mass

Ex 1: How many grams of ammonia are produced from 4.00 grams of hydrogen gas in excess Nitrogen? Balanced Equation: N2 (g) H2 (g)  2NH3 (g) Given : Want : Relationships: 4.00 g H2 ? g NH3 2 mol NH3 = 3 mol H2 1 mol H2 = 2.016g H2 Remember your Steps to Success! 1 mol NH3 = g NH3

Ex 1: How many grams of ammonia are produced from 4 grams of hydrogen gas in excess Nitrogen?
Balanced Equation: N2 (g) H2 (g)  2NH3 (g) Given : Want : Relationships: 4.00 g H2 ? g NH3 2 mol NH3 = 3 mol H2 1 mol H2 = 2.016g H2 1 mol NH3 = g NH3 4.00 g H2 1 mol H2 2 mol NH3 17.031g NH3 x x x 2.016g H2 1 3 mol H2 1 mol NH3 = g NH3 22.5

Ex 2: How many grams of Nitrogen react with 12.0 grams of Hydrogen?
Balanced Equation: N2 (g) H2 (g)  2NH3 (g) Given : Want : Relationships: 12.0 g H2 ? g N2 1 mol H2 = 2.016g H2 Remember your Steps to Success! 1 mol N2 = g N2

Ex 2: How many grams of Nitrogen react with 12.0 grams of Hydrogen?
Balanced Equation: N2 (g) H2 (g)  2NH3 (g) Given : Want : Relationships: 12.0 g H2 ? g N2 1 mol N2 = 3 mol H2 1 mol H2 = 2.016g H2 1 mol N2 = g N2 12.0 g H2 1 mol H2 1 mol N2 28.014g N2 x x x 2.016g H2 1 3 mol H2 1 mol N2 = g N2 55.6

Practice Determine the mass of NH3 produced from 280 g of N2.
What mass of nitrogen is needed to produce 100. kg of ammonia? 340 g NH3 8.22 x 104 g N2

3. MC/F.UN/ Atoms to Mass

Ex 1: How many grams of ammonia are produced from 2
Ex 1: How many grams of ammonia are produced from 2.09 x 1014 mc of hydrogen gas? Balanced Equation: N2 (g) H2 (g)  2NH3 (g) Given : Want : Relationships: 2.09 x 1014 mc H2 ? g NH3 2 mol NH3 = 3 mol H2 1 mol NH3 = g NH3 1 mol H2 = 6.02 x 1023 mc H2

Ex 1: How many grams of ammonia are produced from 2
Ex 1: How many grams of ammonia are produced from 2.09 x 1014 mc of hydrogen gas? Balanced Equation: N2 (g) H2 (g)  2NH3 (g) Given : Want : Relationships: 2.09 x 1014 mc H2 ? g NH3 2 mol NH3 = 3 mol H2 1 mol H2 = 6.02 x 1023 mc H2 1 mol NH3 = g NH3 1 mol H2 2 mol NH3 2.09 x 1014 mc H2 17.031g NH3 x x x 3 mol H2 1 6.02 x 1023 mc H2 1 mol NH3 = g NH3 3.94 x 10-9

Ex 2: How many molecules of ammonia are produced from 40
Ex 2: How many molecules of ammonia are produced from 40.2 g of H2 in the presence of excess N2? Balanced Equation: N2 (g) H2 (g)  2NH3 (g) Given : Want : Relationships: 40.2g H2 ? mc NH3 2 mol NH3 = 3 mol H2 1 mol H2 = 2.016g H2 1 mol NH3 = 6.02 x 1023 mc NH3

Ex 2: How many molecules of ammonia are produced from 40
Ex 2: How many molecules of ammonia are produced from 40.2 g of H2 in the presence of excess N2? Balanced Equation: N2 (g) H2 (g)  2NH3 (g) Given : Want : Relationships: 40.2g H2 ? mc NH3 2 mol NH3 = 3 mol H2 1 mol H2 = 2.016g H2 1 mol NH3 = 6.02 x 1023 mc NH3 40.2g H2 1 mol H2 2 mol NH3 6.02 x 1023 mc NH3 x x x 2.016g H2 3 mol H2 1 1 mol NH3 = mc NH3 8.00 x 1024

Practice How many mc of nitrogen will react with exactly 7.04 grams of hydrogen? How many mc of ammonia will x 1021 mc of N2 produce in excess hydrogen? 7.01 x 1023 mc N2 4.18 x 1021 mc NH3

1) How many mc of nitrogen will react with exactly 7
1) How many mc of nitrogen will react with exactly 7.04 grams of hydrogen? Balanced Equation: N2 (g) H2 (g)  2NH3 (g) Given : Want : Relationships: 7.04 g H2 ? mc N2 1 mol N2 = 3 mol H2 1 mol H2 = 2.016g H2 1 mol N2 = 6.02 x 1023 mc N2 7.04 g H2 1 mol H2 1 mol N2 6.02 x 1023 mc N2 x x x 2.016g H2 3 mol H2 1 1 mol N2 = mc N2 7.01 x 1023

How many mc of ammonia will 2
How many mc of ammonia will x 1021 mc of N2 produce in excess hydrogen? Balanced Equation: N2 (g) H2 (g)  2NH3 (g) Given : Want : Relationships: 2.09 x 1013 mc N2 ? mc NH3 2 mol NH3 = 1 mol N2 1 mol N2 = 6.02 x 1023 mc N2 1 mol NH3 = 6.02 x 1023 NH3 2.09 x 1013 mc N2 1 mol N2 2 mol NH3 6.02 x 1023 mc NH3 x x x 1 1 mol N2 6.02 x 1023 mc N2 1 mol NH3 = mc NH3 4.18 x 1021

4. Molar Volume of Gases (Liters)

4. Molar Volume of Gases (liters)
There is one new conversion: 1 mol of an ideal gas = 22.4 L of an ideal gas. For now we will assume every gas is an ideal gas at STP. You follow the same steps we have been following all along to solve this problem. In a reaction involving gases, Avogadro found that their volumes combine in whole number ratios. In other words, the coefficients in the chemical reaction also represent volume.

For gases: When you see liters…
1 mol __ = 22.4 L __

Ex 1: How many moles of nitrogen gas are needed to react with 44.8 liters of hydrogen gas to produce ammonia gas? Balanced Equation: N2 (g) H2 (g)  2NH3 (g) Given : Want : Relationships: 44.8 L H2 ? mol N2 1 mol N2 = 3 mol H2 1 mol H2 = 22.4 L H2 Remember your Steps to Success! No grams, No molar mass No fun, mc, No 6.02 x 1023!!

44.8 L H2 1 mol H2 1 mol N2 = mol N2 0.667 22.4 L H2 3 mol H2 1
Ex 1: How many moles of nitrogen gas are needed to react with 44.8 liters of hydrogen gas to produce ammonia gas? Balanced Equation: N2 (g) H2 (g)  2NH3 (g) Given : Want : Relationships: 44.8 L H2 ? mol N2 1 mol N2 = 3 mol H2 1 mol H2 = 22.4 L H2 44.8 L H2 1 mol H2 1 mol N2 = mol N2 0.667 x x 22.4 L H2 3 mol H2 1 Given over 1 1 mol = 22.4 L Mole ratio

Ex 2: If 5.00 moles of H2 react with excess N2, how many liters of NH3 are produced? N2 (g) H2 (g)  2NH3 (g) Given : Want : Relationships: 5.00 mol H2 ? L NH3 2 mol NH3 = 3 mol H2 1 mol NH3 = 22.4 L NH3 Remember your Steps to Success! No grams, No molar mass No fun, mc, No 6.02 x 1023!!

Ex 2: If 5.00 moles of H2 react with excess N2, how many liters of NH3 are produced?
N2 (g) H2 (g)  2NH3 (g) Given : Want : Relationships: 5.00 mol H2 ? L NH3 2 mol NH3 = 3 mol H2 1 mol NH3 = 22.4 L NH3 5.00 mol H2 2 mol NH3 22.4 L NH3 = L NH3 74.7 x x 3 mol H2 1 mol NH3 1 Given over 1 Mole ratio 1 mol = 22.4 L

II. Limiting Reactants and Excess Reactants

“which reactant are you going to run out of first?”
In reality, a scientist does not always add chemicals in perfect proportions. There is usually one reactant that is in “excess”. The limiting reactant or limiting reagent is the reactant that limits the amounts of the other reactants that can combine and the amount of product that can form in a chemical reaction. “which reactant are you going to run out of first?”

“which reactant do you have extra of?”
The substance that is not used up completely in a reaction is sometimes called the excess reactant (excess) Once one of the reactants is used up, no more product can be formed “which reactant do you have extra of?”

Which is the limiting reactant
+ 4 4 13 Which is the limiting reactant 4 shells 1 car = 4 cars can be made if 4 shells are available 1 shell 13 tires 1 car = 3.25 cars can be made if 13 tires are available 4 tires Less product can be produced, therefore tires must be the limiting reactant

Activity Write a recipe for the perfect burger

Ex. Consider the following balanced equation: Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g) If 1.21 moles of zinc are added to 2.64 moles of HCl, which reactant is in excess, and which is the limiting reactant? Available Needed 1.21 mol Zn 2 mol HCl = 2.42 mol HCl 1 mol Zn 2.64 mol HCl 1 mol Zn = 1.32 mol Zn 2 mol HCl

< > Limiting reactant Excess reactant Available VS Needed
Ex. Consider the following balanced equation: Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g) If 1.21 moles of zinc are added to 2.64 moles of HCl, which reactant is in excess, and which is the limiting reactant? Available VS Needed < 1.21 mol Zn mol Zn Limiting reactant > 2.64 mol HCl mol HCl Excess reactant

To find the limiting reactant and the excess reactant:
1) Determine the balanced chemical equation 2) Convert the given/available amounts reactants to the number of moles of the other reactant(s) 3) Compare the available amounts to the needed amounts: *If the available amount > needed amount, it is excess reactant * If the available amount < needed amount, it is the limiting reactant 4) To calculate the amount of excess: -Available Amount – Needed Amount = Amount of Excess 5) To calculate the amount of product produced: -G: amount of available limiting reactant -W: amount of product

Example Some rocket engines use a mixture of hydrazine, N2H4, and hydrogen peroxide, as the propellant. The products formed are nitrogen gas and steam. Which is the limiting reactant when 0.75 mol of N2H4 is mixed with 0.50 mol of H2O2? How much excess reactant, in moles, remains unchanged? How much of each product, in moles, is formed?

Step 1: Write a balanced reaction
Some rocket engines use a mixture of hydrazine, N2H4, and hydrogen peroxide, as the propellant. The products formed are nitrogen gas and steam. N2H4 + H2O2 2 N2(g) + H2O(g) 4

N2H4 + H2O2 2 N2(g) + H2O(g) 4 Which is the limiting reactant when mol of N2H4 is mixed with mol of H2O2? Available Needed Limiting Reactant 0.75 mol N2H4 2 mol H2O2 = 1.50 mol H2O2 1 mol N2H4 Excess 0.50 mol H2O2 1 mol N2H4 = 0.25 mol N2H4 2 mol H2O2

To find the limiting reactant and the excess reactant:
4. To calculate the amount of excess: -Available Amount – Needed Amount = Amount of Excess

*Available – Needed = Remaining excess
How much excess reactant, in moles, remains unchanged? Available Needed 0.75 mol N2H4 2 mol H2O = 1.50 mol H2O2 1 mol N2H4 0.50 mol H2O2 1 mol N2H = 0.25 mol N2H4 Excess 2 mol H2O2 - = 0.50 mol N2H4 *Available – Needed = Remaining excess

To find the limiting reactant and the excess reactant:
5. To calculate the amount of product produced: -G: amount of available limiting reactant -W: amount of product

How many moles of product are formed?
N2H4 + H2O2 2 N2(g) + H2O(g) 4 G: W: R: 0.50 mol H2O2 G: amount of available limiting reactant mole N2 N2 is the first product 2 mol H2O2 = 1 mol N2 0.50 mol H2O2 1 mol N2 = 0.25 mol N2 2 mol H2O2 G: W: R: 0.50 mol H2O2 G: amount of available limiting reactant H2O is the second product mole H2O 2 mol H2O2 = 4 mol H2O 0.50 mol H2O2 4 mol H2O = 1.0 mol H2O 2 mol H2O2

Balanced Equation: 8 Zn + S8  8 ZnS
Ex. Zinc metal and solid sulfur (S8) react to form solid zinc sulfide. If 2.0 moles of zinc are heated with 2.0 mol S8, identify the limiting reactant. How many moles of excess reactant remain. How many moles of product are formed? Balanced Equation: 8 Zn + S8  8 ZnS Available Needed 2.0 mol Zn 1 mol S8 = 0.25 mol S8 8 mol Zn 1.0 mol S8 8 mol Zn = 8.0 mol Zn 1 mol S8

< > Limiting reactant Excess reactant Available VS Needed
Ex. Zinc metal and solid sulfur (S8) react to form solid zinc sulfide. If 2.0 moles of zinc are heated with 1.0 mol S8, identify the limiting reactant. How many moles of excess reactant remain. How many moles of product are formed? Available VS Needed < 2.0 mol Zn mol Zn Limiting reactant > 1.0 mol S mol S8 Excess reactant

How much excess is present?
Ex. Zinc metal and solid sulfur (S8) react to form solid zinc sulfide. If 2.0 moles of zinc are heated with 1.0 mol S8, identify the limiting reactant. How many moles of excess reactant remain. How many moles of product are formed? How much excess is present? (Available Amount – Needed Amount = Amount of Excess) 1.0 mol S8 – 0.25 mol S8 = 0.75 mol S8 in excess

To find the limiting reactant and the excess reactant:
4. To calculate the amount of excess: -Available Amount – Needed Amount = Amount of Excess 5. To calculate the amount of product produced: -G: amount of available limiting reactant -W: amount of product

Balanced Equation: 8 Zn + S8  8 ZnS How much product was formed?
Ex. Zinc metal and solid sulfur (S8) react to form solid zinc sulfide. If two moles of zinc are heated with 1.00 mol S8, identify the limiting reactant. How many moles of excess reactant remain. How many moles of product are formed? Balanced Equation: 8 Zn + S8  8 ZnS How much product was formed? G: W: R: 2 mol Zn mol ZnS 8 mol Zn = 8 mol ZnS 2 mol Zn 8 mol ZnS = 2 mol ZnS formed 8 mol Zn

Balanced Equation: C + H2O  H2 + CO
Practice 1. Carbon reacts with steam at high temperatures to produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam, identify the limiting reactant. How many moles of each product are formed? Balanced Equation: C + H2O  H2 + CO Available Needed 2.4 mol C 1 mol H2O = 2.4 mol H2O 1 mol C 3.1 mol H2O 1 mol C = 3.1 mol C 1 mol H2O

< > Limiting reactant Excess reactant Available VS Needed
Practice 1. Carbon reacts with steam at high temperatures to produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam, identify the limiting reactant. How many moles of each product are formed? Available VS Needed < 2.4 mol C mol C Limiting reactant > 3.1 mol H2O mol H2O Excess reactant

To find the limiting reactant and the excess reactant:
4. To calculate the amount of excess: -Available Amount – Needed Amount = Amount of Excess 5. To calculate the amount of product produced: -G: amount of available limiting reactant -W: amount of product

Balanced Equation: C + H2O  H2 + CO
Carbon reacts with steam at high temperatures to produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam, identify the limiting reactant. How many moles of each product are formed? Balanced Equation: C + H2O  H2 + CO G: W: R: 2.4 mol C G: amount of available limiting reactant mole H2 H2 is the first product 1 mol C = 1 mol H2 2.4 mol C 1 mol H2 = 2.4 mol H2 1 mol C G: W: R: 2.4 mol C G: amount of available limiting reactant CO is the second product mole CO 1 mol C = 1 mol CO 2.4 mol C 1 mol CO = 2.4 mol CO 1 mol C

Set VI: Reactions 2) Zn + Pb(NO3)2  Pb + Zn(NO3)2
3) Fe + 2HCl  H2 + FeCl2

III. Theoretical yield, Actual yield and Percent yield

C. Theoretical yield, Actual yield and Percent yield
The yield of a chemical reaction is the quantity of product one obtains from a given ratio of reactants. The actual yield of a chemical reaction is the mass of the compound that you actually recover when you are done with the reaction. The actual yield is also referred to as the experimental yield The theoretical yield is the mass of compound you should obtain (theoretically) if everything goes perfectly. In all of the examples above we have been pretending that everything is perfect. All of our wanteds have been theoretical.

Ex 1: What is the theoretical yield of Na2SO4 in grams if 35 moles of NaOH is reacted with sufficient H2SO4? Yield means product Theoretical yield means mathematically, what should you get? (this is why we do stoichiometry calculations)

Ex 1: What is the theoretical yield of Na2SO4 in grams if 3
Ex 1: What is the theoretical yield of Na2SO4 in grams if 3.50 moles of NaOH is reacted with sufficient H2SO4? 2NaOH + H2SO4  Na2SO4 + 2H2O G: W: R: 3.50 mol NaOH 2 different substances ? g Na2SO4 You see g Na2SO4 2 mol NaOH = 1 mol Na2SO4 1 mol Na2SO4 = g Na2SO4 3.50 mol NaOH 1 mol Na2SO4 g Na2SO4 x x 2 mol NaOH 1 mol Na2SO4 1 Given over 1 Mole ratio 1 mol = g = g Na2SO4 249

If they give you the actual yield and you figure out the theoretical yield, you can find the percent yield. Actual Yield Theoretical Yield X 100 = % Yield Same as… Actual Yield Theoretical Yield % Yield 100 =

Your percent yield should not be greater than 100 because the theoretical yield is the MAXIMUM yield you can have.

Example: From the examples above, if 221 grams of sodium sulfate were actually collected, what is the percent yield? From the experiment 221 g NaSO4 249 g NaSO4 X 100 = 88.8% From the calculation

Practice A student conducts a single displacement reaction that produces 2.75 grams of copper. Mathematically he determines that grams of copper should have been produced. Calculate the student's percentage yield. A student completely reacts 5.00 g of magnesium with an excess of oxygen to produce magnesium oxide. Analysis reveals 8.10 g of magnesium oxide. What is the student's percentage yield? (*Hint: you must first find the theoretical yield.) 87.3% 97.7%

Practice Actual yield: 93.7 g
3. Consider the reaction below of zinc and hydrochloric acid to produce zinc chloride and hydrogen gas. Zn + 2HCl  H2 +ZnCl2 If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of ZnCl2? What is the actual yield in grams if the percent yield is 87%. Actual yield: 93.7 g

Practice 1 A student conducts a single displacement reaction that produces 2.75 grams of copper. Mathematically he determines that 3.15 grams of copper should have been produced. Calculate the student's percentage yield. Actual from the experiment 2.75 g Cu X 100 = 87.3 % 3.15 g Cu calculated

actual X 100 = % yield Theo. Practice 2
A student completely reacts 5.00g of magnesium with an excess of oxygen to produce magnesium oxide. Analysis reveals 8.10 g of magnesium oxide. What is the student's percentage yield? (*Hint: you must first find the theoretical yield.) Actual yield actual X 100 = % yield Theo.

To calculate the theoretical yield
2Mg + O2  2 MgO G: W: R: 5.00 g Mg 2 different substances g MgO You see g Mg and g MgO 2 mol Mg = 2 mol MgO 1 mol Mg = g Mg 1 mol MgO = g MgO 5.00 g Mg 1 mol Mg 2 mol MgO g MgO x x x g Mg 2 mol Mg 1 1 mol MgO Theoretical yield= g MgO 8.29

Practice 2 A student completely reacts 5.00g of magnesium with an excess of oxygen to produce magnesium oxide. Analysis reveals 8.10 g of magnesium oxide. What is the student's percentage yield? (*Hint: you must first find the theoretical yield.) Actual yield 8.10 g MgO X 100 = 97.7% 8.29 g MgO

Practice 3 Actual yield: 93.7 g
3. Consider the reaction below of zinc and hydrochloric acid to produce zinc chloride and hydrogen gas. Zn + 2HCl  H2 +ZnCl2 If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of ZnCl2? What is the actual yield in grams if the percent yield is 87%. Actual yield: 93.7 g

G: 58.0 g HCl W: R: g ZnCl2 Zn + 2HCl  H2 +ZnCl2
If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of ZnCl2? What is the actual yield in grams if the percent yield is 87%. Zn + 2HCl  H2 +ZnCl2 G: W: R: 58.0 g HCl 2 different substances g ZnCl2 You see g HCl & g ZnCl2 2 mol HCl = 1 mol ZnCl2 1 mol HCl = g HCl 1 mol ZnCl2 = g ZnCl2 58.0 g HCl 1 mol HCl 1 mol ZnCl2 g ZnCl2 x x x g HCl 2 mol HCl 1 1 mol ZnCl2 Theoretical yield= g ZnCl2 108

Practice 3 If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of ZnCl2? What is the actual yield in grams if the percent yield is 87%. Actual from the experiment 93.7 g actual X 100 = 87 % 108 g calculated

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