Drill – Empirical/Molecular Formulas Drill

• Please complete #1 and #2

Agenda

• Pass fwd empirical/molecular form WS

• Limiting Reactants WS

• Reminder: Mole Project due Mar 18

• Stoichiometry Test - Friday

Limiting Reactants

ExampleYour job is to attach arms on dolls in a doll factory. If you have 600 arms and 350 dolls, how many finished dolls can you make?

What do you havein excess?

What is limiting your production – the arms or the dolls?

• Reagent = reactant

• Limiting Reactant – the reactant that is completely consumed. This reactant limits the amount of product that is formed.

• Excess Reactant – the reactant that is not limiting the reaction. There will be leftovers of this reactant.

How to solve for limiting reactant

1. Write the balanced chemical equation

2. Determine the moles of each reactant

3. Determine how many moles of product each reactant would make using a mole ratio

4. The reactant that yields less product is the limiting reactant.

• Start here pd 3

1. Write the balanced chemical equation

Copper reacts with sulfur to form copper (I) sulfide. What is the limiting reagent when 80.0g Cu reacts with 25.0g S?

Cu S+ Cu2S2

2. Determine the moles of each reactant.

Copper reacts with sulfur to form copper (I) sulfide. What is the limiting reactant when 80.0g Cu reacts with 25.0g S?

80.0 g Cu x

25.0 g S x

1 mole Cu

1 mole S

63.55 g Cu

32.06 g S

= 1.26 mol Cu

= .780 mol S

3. Determine how many moles of product each reactant would make using a mole ratio

Copper reacts with sulfur to form copper (I) sulfide. What is the limiting reactant when 80.0g Cu reacts with 25.0g S?

1.26 mol Cu x

.780 mol S x

1 mol Cu2S2 mol Cu

1 mol S

= .630 mol Cu2S

= .780 mol Cu2S1 mol Cu2S

4. The reactant that yields less product is the limiting reagent.

Copper reacts with sulfur to form copper (I) sulfide. What is the limiting reactant when 80.0g Cu reacts with 25.0g S?

Copper yields .630 mol Cu2S

Sulfur yields .780 mol Cu2S

So Copper is the limiting reagent

Amount of Excess?

• In order to determine how much (mass) of excess reactants there are, you must figure out what mass of the excess reactant got used up…then subtract that from the initial mass.

Amount of Excess?• Step 1 – Start with moles of product

formed (from limiting reactant).

• Step 2 – Convert back to moles of excess reactant.

• Step 3 – Convert from moles to mass of excess reactant.

• Step 4 – Subtract this value from the original mass of excess reactant (given in problem)

• Start here – pd 4B

Sample Problem

3 Fe + 4 H2O Fe3O4 + 4H2

– When 36.0 g of water is mixed with 67.0 g of Fe, which is the limiting reactant?

– Determine the grams of iron oxide produced.– Determine the mass of excess reactant

remaining.

1. Convert both masses to moles and calculate the number of moles of Fe3O4

67.0 g Fe = 0.400 mol Fe3O4

36.0 g H20 = 0.499 mol Fe3O4

2. Convert back to mass of excess reactant

0.400 mol Fe3O4 x 4 mol H20 x 18.02 g H20 =

1 mol Fe3O4 1 mol H20

28.8 g H20

3. Subtract this value from the original mass of excess reagent (given in problem)

36.0 g H2O – 28.8 g H2O consumed =

7.2 g H2O remaining

Homework

• Mole Project

• Limiting Reactant Worksheet

Pd 4a – start here

Agenda

• Stoichiometry Test

• Hydrates with Worksheet

• Stochiometry Lab Data

Hydrates

• When salts combine chemically with water and form unstable hydrated crystals

• Although in a chemical bond, this water of hydration maintains its characteristic composition

• Instead of H20Na2SO14, we write

Na2SO4 H2O

Homework

• Hydrates WS

• Stoichiometry Test

Drill – pd 3

• Determine the percentage of water in the following hydrate:

Sc(NO3)3 · 10H2O [scandium(III) nitrate decahydrate]

Answer

• 43.8% water

Drill – all classes

In an experiment, 3.25 g of NH3 are allowed to react with 3.50 g of O2.

NH3 + O2 NO + H2O

a. Which reactant is the limiting reagent?  

b. How many grams of NO are formed?  

c. How much of the excess reactant remains after the reaction?  

Drill – all classes

In an experiment, 3.25 g of NH3 are allowed to react with 3.50 g of O2.

NH3 + O2 NO + H2O

a. Which reactant is the limiting reagent?  O2

b. How many grams of NO are formed?  2.63 g NO

c. How much of the excess reactant remains after the reaction?  1.71 g NH3 left

Make sure to turn in :

• Composition of Hydrates WS

• Limiting Reactants WS

• Lab Calculations

Next class

• Wear close-toed shoes – Lab

• Wed or Th will be your test

• Does anyone need a mole pattern?

Agenda

• Lab Calculations

• Mixed Stoichiometry/Limiting Reactants WS

• Review Sheet if finished

Drill

• Read the Introduction of Lab

• Change Lab to read Percent Composition of Epsom Salt

Lab

• Only heat crucible once for 10 minutes

• Allow it to cool for 10 minutes

• Make sure crucible is cool before handling it.

Lab Groups – pd 4B1. Stephanie, Chris H.

2. Ben, Comfort

3. Jessica, Zach

4. John, Chris J

5. Sahra, O

6. Amelia, Valentina

7. Emileigh, Caine

8. Troy, Claire,Max

9. Shivani, Anjali

10.Josh, Becky

11.Sophia, Max

12.Joey, Alexis

Homework

• Review WS