# Limiting Reactant Percent Yield. Consider the following reaction 2 H 2 + O 2  2 H 2 O.

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Limiting Reactant Percent Yield

Consider the following reaction 2 H 2 + O 2  2 H 2 O

Reactants are combined in perfect proportions 6 molecules 3 molecules

This is a theoretic reaction 6 molecules 3 molecules

In reality this never happens 6 molecules 3 molecules

Consider 4 molecules 4 molecules + leftover oxygen 3 molecules

Consider LIMITING REACTANT Amount of PRODUCT is determined by limiting reactant EXCESS REACTANT

Consider 6 molecules 4 molecules + leftover hydrogen 2 molecules

Consider EXCESS REACTANT Amount of PRODUCT is determined by limiting reactant LIMITING REACTANT

Given 24 grams of O 2 and 5.0 grams of H 2 determine the mass of H 2 O produced. 2 H 2 + O 2  2 H 2 O the mass of H 2 O produced will be determined by the limiting reactant - do TWO calculations

calculation for 24 grams of O 2 24 g O 2 2 H 2 O18.0 g mol -1 32.0 g mol -1 1 O 2 = 27 g of H 2 O

calculation for 24 grams of O 2 24 g O 2 2 H 2 O18.0 g mol -1 32.0 g mol -1 1 O 2 = 27 g of H 2 O calculation for 5.0 grams of H 2 5 g H 2 2 H 2 O18.0 g mol -1 2.0 g mol -1 2 H 2 = 45 g of H 2 O

calculation for 24 grams of O 2 24 g O 2 2 H 2 O18.0 g mol -1 32.0 g mol -1 1 O 2 = 27 g of H 2 O calculation for 5.0 grams of H 2 5 g H 2 2 H 2 O18.0 g mol -1 2.0 g mol -1 2 H 2 = 45 g of H 2 O O 2 is the LIMITING REACTANT and determines the amount of product H 2 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 oxygen given 24 grams of O 2 24 g O 2 2 H 2 2.0 g mol -1 32.0 g mol -1 1 O 2 = 3.0 g of H 2 3.0 g of H 2 reacts so

How much hydrogen gas would be left over? To calculate, first determine how much reacts with all of the oxygen given 24 grams of O 2 24 g O 2 2 H 2 2.0 g mol -1 32.0 g mol -1 1 O 2 = 3.0 g of H 2 3.0 g of H 2 reacts so 5.0 g – 3.0 g = 2.0 g of hydrogen remains

Percent Yield 2 AuCl 3 +3 Pb  3 PbCl 2 + 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 AuCl 3 +3 Pb  3 PbCl 2 + 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 AuCl 3 +3 Pb  3 PbCl 2 + 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 AuCl 3 +3 Pb  3 PbCl 2 + 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 AuCl 3 14.0 g AuCl 3 2 AuCl 3 196.97 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 AuCl 3 is not pure

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

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