Ch. 3.9-3.10: Limiting Reagents; Reaction Yield Name • Limiting Reagent is used up first in a reaction & controls the amount of product that can be

produced; Excess Reagent is present in greater quantity than needed and has an amount remaining

• How to identify the limiting reagent for general reaction aA + bB ® cC + dD, given the moles of the reactants: 1. Compare moles of reaction from A & B:

§ ; Smaller mol Rxn is limiting reagent;

• If they are equal there is no limiting reagent and both are used up 2. To determine amount of the excess reagent, determine the amount of the excess reagent

needed to fully react with the limiting reagent and subtract the needed amount of excess reagent from given amount of excess reagent.

• For the reaction 5 O2 (g) + 4 NH3 (g) → 4 NO (g) + 6 H2O (g), determine limiting reagent if you start with 8.96 g of O2 and 4.09 g of NH3, and the amount of excess reagent remaining:

O2 is limiting since 0.0570 mol Rxn < 0.0600 mol Rxn; thus NH3 is in excess • Now find mass of NH3 needed, subtract it from the starting mass to find the excess remaining:

o mass NH3 remaining = 4.09 g – g =

To help organize and visualize this method, look at this ICF table and picture:

N2 (g) + 3 H2 (g) → 2 NH3 (g) Initial: 5 mc 9 mc 0 mc

Change:

Final:

Reaction Yields • Theoretical Yield is the amount of product that would be produced if all of the limiting reagent

reacted and it was 100% pure. o Find from straight mass-mass stoichiometry problem from limiting reagent o For the problem above, the theoretical yield of H2O is:

mol RxnA = mol A × 1 mol Rxna mol A

mol RxnO2= 8.96 g O2 ×

1 mol O2

32.00 g O2× 1 mol Rxn

5 mol O2=

mol RxnNH3= 4.09 g NH3 ×

1 mol NH3

17.04 g NH3× 1 mol Rxn

4 mol NH3=

mass NH3 needed = 0.0560 mol Rxn × 4 mol NH3

1 mol Rxn× 17.04 g NH3

1 mol NH3=

? g H2O = 0.0560 mol Rxn × 6 mol H2O1 mol Rxn

× 18.02 g H2O1 mol H2O

=

• Actual Yield is the amount of product actually obtained in a reaction o Value usually determined experimentally

• Percent Yield is the ratio of actual yield to theoretical yield, multiplied by 100:

o

o Measures the efficiency of a reaction o % yield is usually less than 100% because of:

§ Reversible reactions § Side reactions § Difficulty recovering product § Impurity of reactant

• I.e. a reaction with 75% yield can imply a 75% pure reactant

o What is the percent yield of H2O above if the actual yield is 4.83 g H2O?

o Can use % yield to determine theoretical yield and purity of a reactant: § What was the purity of a 19.8 g sample of impure NH3 in the reaction above (assuming

NH3 is limiting) if one obtains 25.8 g NO?

o Can determine needed mass of reactant for a given mass of product if the % yield is known: § What mass of O2 is needed in the reaction above (assuming O2 is limiting) to produce

45.0 g NO is the % yield is 65.0%?

% yield = actual yield

theoretical yield × 100

% yield = 4.83 g H2O6.05 g H2O

×100 =

? g NH3 = 25.8 g NO × 1 mol NO30.01 g NO

× 4 mol NH3

4 mol NO× 17.04 g NH3

1 mol NH3=

% Purity = g NH3( pure)19.8 g NH3(impure)

×100% =

? g O2 = 45.0 g NO × 100%65.0%

Theoretical Yield = 69.23 g O2

! "############× 1 mol NO

30.01 g NO× 5 mol O2

4 mol NO× 32.00g O2

1 mol O2

Stoichiometry! "#########################

=