2011 topic 01 lecture 3 - limiting reactant and percent yield
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Transcript of 2011 topic 01 lecture 3 - limiting reactant and percent yield
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Lecture 3
Limiting Reactant
Percent Yield
IB Chemistry Power Points
Topic 1
Quantitative Chemistrywww.pedagogics.ca
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Consider the following reaction
2 H2 + O2 2 H2O
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Reactants are combined in perfect proportions
6 molecules 6 molecules3 molecules
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In reality this never happens
6 molecules 6 molecules3 molecules
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Consider
4 molecules 4 molecules +leftover oxygen
3 molecules
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Consider
LIMITING REACTANT
Amount ofPRODUCT is determined by limiting reactant
EXCESS REACTANT
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Consider
6 molecules 4 molecules +leftover hydrogen
2 molecules
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Consider
EXCESS REACTANT
Amount ofPRODUCT is determined by limiting reactant
LIMITING REACTANT
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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
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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
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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 H2
5 g H2 2 H2O 18.0 g mol-1
2.0 g mol-1 2 H2
= 45 g of H2O
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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 H2
5 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)
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How much hydrogen gas would be left over? To calculate, first determine how much reacts with all of the oxygen
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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
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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
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Percent Yield
2 AuCl3 +3 Pb 3 PbCl2 + 2 Au
Enoch the Red, 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:
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Percent Yield
2 AuCl3 +3 Pb 3 PbCl2 + 2 Au
Enoch reacts 14.0 g of gold III chloride with excess lead metal. What would be the maximum, THEORETICAL yield of this reaction?
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Percent Yield
2 AuCl3 +3 Pb 3 PbCl2 + 2 Au
Enoch reacts 14.0 g of gold III chloride with excess lead metal. What would be the maximum, THEORETICAL yield of this reaction?
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Percent Yield
2 AuCl3 +3 Pb 3 PbCl2 + 2 Au
Enoch 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 AuCl3196.97 g mol-1
303.5 g mol-1 2 Au = 9.09 g Au
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Percent Yield Enoch 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
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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