• Announcement: – Term 2 Cut Off Date– Upcoming Assessment before Spring Break

• HI #10 (Key Posted Online in Weekend)• Limiting Reagent

Today’s Plan

When 15.0g of CH4 is reacted with unlimited amount of Cl2 according to:

CH4 + Cl2→ CH3Cl + HCl

What is the mass of CH3Cl produced?MM(CH4)=16.0g/molMM(CH3Cl)= 50.5g/mol

Calculate expected mass of product

50.5g CH3Cl

1 mol CH3Cl

1 mol CH4

16.0 g CH4

1 mol CH3Cl

1 mol CH4

CH4 +Cl2→ CH3Cl + HClMass of CH3Cl produced =

Calculate expected mass of product

15.0g CH4

= 47.3g CH3Cl

MM(CH4)=16.0g/molMM(CH3Cl)= 50.5g/mol

× × ×

When 15.0g of CH4 is reacted with 10.0g of Cl2according to:

CH4 + Cl2→ CH3Cl +HCl

What is the mass of CH3Cl produced?

Limiting reactant calculation

Making Sandwiches!

2 breads + 1 ham à 1 sandwich

You have:

10 slices of bread

3 ham

Q: When 10 slices of bread and 3 ham are

available, how many complete sandwiches

can you make? (Modification to the

ingredient is not allowed!)

Recipe is…

1 SW

2 breads

2 breads + 1 ham → 1 sandwich# of sandwich produced =

How many SW can be made?

10 breads = 5 SW×

3 ham = 3 SW×1 SW

1 ham

The # of ham limits the amount of sandwich.

Bread Ham

Have 10 3

# SW I can make 5 3

• We could potentially make 5 sandwiches (we have 10 slices of bread), but the # of ham limits the # of sandwich we can make…

• Ham limits the amount of product à Ham is the “limiting reactant” (LR)

• As soon as the ham runs out, we cannot make anymore sandwich and there will be some bread left over.

• Excess bread remains à Bread is the “excess reactant” (ER)

2 breads + 1 ham à 1 sandwich

When 15.0g of CH4is reacted with 10.0g of Cl2according to:

CH4+Cl2→ CH3Cl + HCl

What is the mass of CH3Cl produced?1. 15.0 g CH4 à CH4 mol à CH3Cl mol à CH3Cl

g2. 10.0g Cl2 à Cl2 mol à CH3Cl mol à CH3Cl g

Limiting reactant calculation

50.5g CH3Cl

1 mol CH3Cl

1 mol Cl271.0 g Cl2

1 mol CH3Cl

1 mol Cl2

CH4+Cl2→ CH3Cl + HClMass of CH3Cl produced =

Calculate expected mass of product

10.0g Cl2

=7.11g CH3Cl

× × ×

MM(Cl2)=71.0g/mol

MM(CH3Cl)= 50.5g/mol

Using CH4 Using Cl2

CH3Clproduced

47.3 g 7.11g

• There is enough CH4 to make 47.3g of CH3Cl, but only enough Cl2 to make 7.11g of CH3Cl. Cl2 sets a limit. Cl2 is called the LIMITING REAGENT (LR).

• On the other hand, CH4 is called the EXCESS REAGENT (ER) because not all of it gets used and there will be some left over.

CH4+Cl2→ CH3Cl+HCl

Calculate expected mass of product

Using CH4 Using Cl2

CH3Clcalculated

47.3 g 7.11g

Cl2 is the Limiting Reactant

The smallest amount of CH3Cl calculated (7.11 g is known as the MAXIMUM YIELD of product or the THEORETICAL YIELD.

CH4 is the Excess Reactant

• What calculation did we do?– Calculate how much a product* could be made

based on each reactant, then compare the results. (2x stoichiometry)

– The reactant that makes the least amount of product is the LR.

*When there are multiple products, pick one and stick to that throughout the calculation.

Limiting Reactant (LR)

18.0g H2O

1 mol H2O

1 mol H2

2.0 g H2

2 mol H2O

2 mol H2

2H2+O2→ 2H2OMass of H2O produced (From H2)=

Quick Practice

10.0g H2

= 90.g H2O

MM(H2)=2.0g/molMM(O2)=32.0g/molMM(H2O)= 18.0g/mol

× × ×

18.0g H2O

1 mol H2O

1 mol O2

32.0 g O2

2 mol H2O

1 mol O2

Mass of H2O produced (From O2)=

50.0g O2

= 56.3g H2O

× × ×

O2 is the LR and H2 is the ER

• Study for Exo/Endothermic Quiz• “Limiting Reagent and Percent Yield” w/s in HB

Question 1-7• PreLab 6B Due March 1st (Fri) Day 1 and March

4th (Mon) Day 2

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