LIMITING REACTANT

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Chemistry 101 Chapter 3
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LIMITING REACTANT
• When 2 or more reactants are combined in non-stoichiometric ratios, the amount of product produced is limited by the reactant that is not in excess (limiting reactant). Analogy: The number of sundaes possible is limited by the amount of syrup, the limiting reactant.

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Chemistry 101 Chapter 3
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Limiting Reactant (Reagent) Problems always involve 2 steps: 1.
Identify the Limiting Reactant (LR)
convert all masses to moles compare actual mole ratio to mole ratio given by the the balanced chemical equation OR calculate the number of moles obtained from each reactant in turn. The reactant that gives the smaller amount of product is the Limiting R.eactant. 2.
Calculate the amount of product obtained from the Limiting Reactant
Example 1 Sodium hydrogen carbonate is prepared from NaCl and ammonium hydrogen carbonate, according to the equation:
NH4HCO3(aq) + NaCl(aq) NaHCO3(aq) + NH4Cl(aq)
If 0.300 moles of NH4HCO3 are reacted with 0.2567 moles of NaCl, how many grams of NaHCO3 are obtained ?
1 NH4HCO3(aq) + 1 NaCl(aq) 1 NaHCO3(aq) + 1 NH4Cl(aq)
0.300 moles 0.2567 moles ? g
L.R. 1 mole NaHCO3
84.01 g NaHCO3 ? g NaHCO3 = 0.2567 moles NaCl x ------- x ------
1 mole NaCl 1 mole NaHCO3
= 21.57 g NaHCO3

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Chemistry 101 Chapter 3
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Example 2 A 1.4 g sample of magnesium is treated with 8.1 g of hydrochloric acid to produce magnesium chloride and hydrogen gas. How many grams of hydrogen are produced ?
Mg(s) +
2 HCl(aq)
MgCl2(aq) + H2(g) 1.4 g 8.1 g
? g
Change masses of reactants in moles:
1 Mg(s)
+ 2 HCl(aq)
1 MgCl2(aq) + 1 H2(g)
1 mole 1 mole 1.4 g x --- 8.1 g x --- 24.31 g 36.46 g 0.0576 moles 0.222 moles
L.R. requires 2 x 0.0576 moles HCl = 0.115 moles HCl (0.222 moles HCl) available
HCl is an excess!
1 mole H2 2.02 g H2 ? g H2 = 0.0576 moles Mg x ----- x ----- = 0.12 g H2 1 mole Mg 1 mole H2
L.R.

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Chemistry 101 Chapter 3
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Solution recommended by textbook:
1 Mg(s)
+ 2 HCl(aq)
1 MgCl2(aq) + 1 H2(g)
1 mole 1 mole 1.4 g x --- 8.1 g x --- 24.31 g 36.46 g 0.0576 moles 0.222 moles
Calculate the number of moles obtained from each reactant in turn. The reactant that gives the smaller amount of product is the Limiting Reactant.
1 mole H2 2.02 g H2 ? g H2 = 0.222 moles HCl x ----- x ----- = 0.22 g H2 2 moles HCl 1 mole H2 1 mole H2 2.02 g H2 ? g H2 = 0.0576 moles Mg x ----- x ----- = 0.12 g H2 1 mole Mg 1 mole H2 smaller ! (correct answer) Since Mg produces the smaller amount of product, Mg is the L.R.

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Chemistry 101 Chapter 3
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THE YIELD CONCEPT
• Quantities of product calculated represent the maximum amount obtainable (100 % yield) • Most chemical reactions do not give 100 % yield of product because of: side reactions (unwanted reactions) reversible reactions ( reactants products) losses in handling and transferring
Actual Yield Percent Yield = -------- x 100 Theoretical Yield
Actual Yield: Amount of product actually obtained (experimental) Theoretical Yield: Maximum amount of product obtainable (calculated from equation)

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Chemistry 101 Chapter 3
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Example 1 A 35.0 g sample of calcium hydroxide is reacted with excess phosphoric acid, according to the following balanced chemical equation:
3 Ca(OH)2(aq)
+ 2 H3PO4(aq)
1 Ca3(PO4)2(s) +
6 H2O(l)
(a) How many grams of calcium phosphate can be produced ?
1 mole Ca(OH)2
1 mole Ca3(PO4)2 310.3 g Ca3(PO4)2
? g Ca3(PO4)2 = 35.0 g Ca(OH)2 x ------- x ------- x ------- = 48.9 g Ca3(PO4)2 74.10 g Ca(OH)2
3 mole Ca(OH)2
1mole Ca3(PO4)2
(b) If 45.2 grams of calcium phosphate are actually obtained in a laboratory experiment, what is the percent yield ?
Actual Yield 45.2 g Percent Yield = -------- x 100 = --- x 100 = 92.4 % Theoretical Yield 48.9 g

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Chemistry 101 Chapter 3
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Example 2 Sodium hydrogen carbonate is prepared from NaCl and ammonium hydrogen carbonate, according to the equation:
NH4HCO3(aq) + NaCl(aq) NaHCO3(aq) + NH4Cl(aq)
If 0.300 moles of NH4HCO3 are reacted with 0.2567 moles of NaCl, and 10.45 g of NaHCO3 are obtained, what is the percent yield? 1. First calculate the maximum amount obtainable (theoretical yield) from the given quantities (theoretical yield)
1 NH4HCO3(aq) +1 NaCl(aq) 1 NaHCO3(aq) + 1 NH4Cl(aq)
0.300 moles 0.2567 moles ? g
L.R. 1 mole NaHCO3
84.01 g NaHCO3 ? g NaHCO3 = 0.2567 moles NaCl x ------- x ------ = 21.57 g NaHCO3 (theoretical yield)
1 mole NaCl
1 mole NaHCO3
2. Second, calculate % yield from actual and theoretical yield
Actual Yield 10.45 g Percent Yield = -------- x 100 = --- x 100 = 48.45 % Theoretical Yield 21.57 g