limiting reactant percent yield

Limiting ReactantPercent Yield

Consider the following reaction

2 H2 + O2 2 H2O

Reactants are combined in perfect proportions

6 molecules 6 molecules3 molecules

This is a theoretic reaction

6 molecules 6 molecules3 molecules

In reality this never happens

6 molecules 6 molecules3 molecules

Consider

4 molecules 4 molecules +leftover oxygen

3 molecules

Consider

LIMITING REACTANT

Amount ofPRODUCT is determined by limiting reactant

EXCESS REACTANT

Consider

6 molecules 4 molecules +leftover hydrogen

2 molecules

Consider

EXCESS REACTANT

Amount ofPRODUCT is determined by limiting reactant

LIMITING REACTANT

Given 24 grams of O2 and 5.0 grams of H2 determine the mass of H2O produced.

2 H2 + O2 2 H2O

the mass of H2O produced will be determined by the

limiting reactant - do TWO calculations

calculation for 24 grams of O2

24 g O2 2 H2O 18.0 g mol-1

32.0 g mol-1 1 O2 = 27 g of H2O

calculation for 24 grams of O2

24 g O2 2 H2O 18.0 g mol-1

32.0 g mol-1 1 O2 = 27 g of H2O

calculation for 5.0 grams of H25 g H2 2 H2O 18.0 g mol-1

2.0 g mol-1 2 H2

= 45 g of H2O

calculation for 24 grams of O2

24 g O2 2 H2O 18.0 g mol-1

32.0 g mol-1 1 O2 = 27 g of H2O

calculation for 5.0 grams of H25 g H2 2 H2O 18.0 g mol-1

2.0 g mol-1 2 H2

= 45 g of H2O

O2 is the LIMITING REACTANT and determines the amount of product

H2 is the EXCESS REACTANT (some would be left over)

How much hydrogen gas would be left over? To calculate, first determine how much reacts with all of the oxygen

How much hydrogen gas would be left over? To calculate, first determine how much reacts with all of the oxygengiven 24 grams of O2

24 g O2 2 H2 2.0 g mol-1

32.0 g mol-1 1 O2 = 3.0 g of H2

3.0 g of H2 reacts so

How much hydrogen gas would be left over? To calculate, first determine how much reacts with all of the oxygengiven 24 grams of O2

24 g O2 2 H2 2.0 g mol-1

32.0 g mol-1 1 O2 = 3.0 g of H2

3.0 g of H2 reacts so

5.0 g – 3.0 g = 2.0 g of hydrogen remains

Percent Yield

2 AuCl3 +3 Pb 3 PbCl2 + 2 Au

Dougie, an alchemist, wants to try to turn lead into gold (which you can’t do chemically). He finds that mixing lead with an unidentified compound (gold III chloride) actually produces small amounts of gold. The reaction is as follows:

Percent Yield

2 AuCl3 +3 Pb 3 PbCl2 + 2 Au

Dougie reacts 14.0 g of gold III chloride with excess lead metal. What would be the maximum, THEORETICAL yield of this reaction?

Percent Yield

2 AuCl3 +3 Pb 3 PbCl2 + 2 Au

Dougie reacts 14.0 g of gold III chloride with excess lead metal. What would be the maximum, THEORETICAL yield of this reaction?

Percent Yield

2 AuCl3 +3 Pb 3 PbCl2 + 2 Au

Dougie reacts 14.0 g of gold III chloride with excess lead metal. What would be the maximum, THEORETICAL yield of this reaction?

given 14.0 g of AuCl3

14.0 g AuCl3 2 AuCl3 196.97 g mol-1

303.5 g mol-1 2 Au = 9.09 g Au

Percent Yield Dougie recovers only 1.05 g of gold from the reaction. This could be for many different reasons

some product was lost in the recovery process

the reaction did not go to completion

the AuCl3 is not pure

Percent Yield The percentage yield expresses the proportion of the expected product that was actually obtained.

actual% yield= ×100%theoretical1.05% yield= ×100%=11.6%9.09