Limiting ReagentsYou know what happens when you don’t have enough

oxygen right?

Solve this:Given: 4NH3 + 5O2 ! 6H2O + 4NO

How many grams of NO are produced if 20.0g of NH3 are burned in 30.0g of O2?

AssumptionsGiven: 4NH3 + 5O2 ! 6H2O + 4NO

How many grams of NO can be made from 20.0g of NH3?

What assumption is being made here?

We are assuming that we have exactly the right amount of oxygen (or an excess) for this reaction

to go to completion

AssumptionsGiven: 6 CH4(g) + 24 O2(g) ! 6 CO2(g) + 12 H2O(g)

What happens if our assumption of excess oxygen is wrong?

6 CH4(g) + 9 O2(g) ! 2 CO2(g) + 2 CO(g) + 2 C(s) + 12 H2O(g)

Incomplete combustion...possible death?

Limiting Reagents - no assumptions

Calculating the quantity of product based on which of the two reactants is limited

For example...Given: 4NH3 + 5O2 ! 6H2O + 4NO

How many moles of NO are produced if:

4 mol NH3 & 5 mol O2? 4 mol NO

4 mol NH3 & 20 mol O2?

8 mol NH3 & 20 mol O2?

4 mol NO O2 in excess

8 mol NO O2 in excess

In the second two, NH3 is the limiting reagent. If we used more NH3 we get more NO.

for example (con’t)Given: 4NH3 + 5O2 ! 6H2O + 4NO

How many moles of NO are produced if:

4 mol NH3 & 2.5 mol O2? 2 mol NO

The quantity of product is limited by the limiting reagent, therefore we use it for our calculations, not

the reactant that is present in excess.

NH3 in excess

Solve this:Given: 4NH3 + 5O2 ! 6H2O + 4NO

How many grams of NO are produced if 20.0g of NH3 are burned in 30.0g of O2?

Step 1 (mass to moles...for each reactant now)

1 mol NH3 17.0 g NH3

x # mol NH3=20.0 g NH3 1.18 mol NH3=

1 mol O2 32.0 g O2

x # mol O2= 30.0 g O20.938 mol O2=

Solve this:Given: 4NH3 + 5O2 ! 6H2O + 4NO

Step 2 (look at your mole ratio & determine the limiting reagent)

1.18 mol NH3 0.938 mol O2:

0.938 mol0.938 mol

1.25 NH3 : 1 O2

This is what you have!!!

4 NH3 : 5 O2This is what you

need!!!

You don’t have enough O2

Solve this:Given: 4NH3 + 5O2 ! 6H2O + 4NO

Step 3 (stoichiometry, using the limiting value for the LR)

30.0g(O2)

5 mol(O2)=0.938 mol(O2)

4 mol(NO)x32.00g/mol(O2)

=0.750 mol(NO)

0.750 mol(NO)x30.01g/mol(NO) = 22.5g(NO)

Oxygen - the Limiting Reagent

This is how much NO you can produce from 20.0g of NH3 & 30.0g of O2

The other methodGiven: 4NH3 + 5O2 ! 6H2O + 4NO

One Step, 2 Calculations

30.0g(O2)

5 mol(O2)=0.938 mol(O2)

4 mol(NO)x32.00g/mol(O2)

=0.750 mol(NO)

0.750 mol(NO) x30.01g/mol(NO) =22.5g(NO)

20.0g(NH3)

4 mol(NH3)= 1.17 mol(NH3)

4 mol(NO)x17.04g/mol(NH3)

= 1.17 mol(NO)

1.17 mol(NO) x30.01g/mol(NO) = 35.1g(NO)

Which one produces the least amount of product?

22.5g(NO)

Homework:

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