Stoichiometry,Limiting Reactants and Percent Yield

• How many grams of carbon dioxide will be produced when 11.6 g of butane (C4H10) burns with oxygen?

2 C4H10 + 13 O2 8 CO2 + 10 H2O

11.6 g ? g

2 mole 8 mole

11.6 g C4H10 1 mole C4H10

X -------------------- 58.1 g C4H10

8 mole CO2 X----------------2 mole C4H10

44.0 g CO2

X --------------- = 1 mole CO2

35.1 g CO2

Excess/Limiting Reactants

• The limiting reactant is the reactant present in the smallest stoichiometric amount– In other words, it’s the reactant you’ll run out of first!

• The excess reactant will be left over after the reaction.

Limiting Reactants Example

a) 11.6 grams of magnesium react with 54.6 grams of nitric acid. Hydrogen and magnesium nitrate are produced. How many grams of hydrogen are produced?

11.6 g 54.6 g x g Mg + 2 HNO3 ----> H2 + Mg(NO3)2

Method 1Here are the 2 possible calculations (you really only need to do one of these):  1 mol Mg 2 mol HNO3 63.0 g HNO3

11.6 g Mg X ----------- X --------------- X -------------- = 60.1 g HNO3

24.3 g Mg 1 mole Mg 1 mole HNO3

We can’t use all the Mg – there’s not enough HNO3 to reactOR

  1 mole HNO3 1 mole Mg 24.3 g Mg

54.6 g HNO3 X-------------- X -------------- X ----------- = 10.5 g Mg

63.0 g HNO3 mole HNO3 mole Mg

If we use all the HNO3 we’ll still have Mg left over

Method 1 cont.

ALWAYS USE THE LIMITING REACTANT TO FIND THE REQUIRED!

  

1 mole HNO3 1 mole H2 2.02 g H2

54.6 g HNO3 X --------------- X --------------- X ---------- = .875 g H2

63.0 g HNO3 2 mole HNO3 1 mole H2

Method 2

Mg 3.24

Mg mol 1*

g

Mg g 6.11

3

2

HNO mol 2

H mol 1*

2

2

H mol 1

H g 2.02* 2H g .8753HNO g 4.65

3

3

HNO g 0.63

HNO mol 1*

20.964 g H2

2

2.02 g H*

1 mol H21 mol H

*1 mol Mg

AND

The smaller answer is correct – 0.875g of H2 is formed.HNO3 is the limiting reactant and Mg is the excess reactantWe know this because we used HNO3 to find the correct answer

Part b…b. How much magnesium is left over after the reaction?

(you may have already done the first step if you happened to do this calculation to find the limiting reactant in method 1)

1 mole HNO3 1 mole Mg 24.3 g Mg

54.6 g HNO3 X-------------- X -------------- X ----------- = 10.5 g Mg

63.0 g HNO3 mole HNO3 mole Mg

11.6g Mg – 10.5 g Mg = 1.1 g Mg is left over!

Theoretical Yield

• The theoretical yield is the amount of product that can be made– This is what we calculate

• The actual yield is the amount one actually produces and measures– Either measured in the lab or given in the

problem

Percent Yield

A comparison of the amount actually obtained to the amount it was possible to make

Actual YieldTheoretical YieldPercent Yield = x 100%