basic chemistry copyright © 2011 pearson education, inc. 1 chapter 9 chemical quantities in...
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Basic Chemistry Copyright © 2011 Pearson Education, Inc.1
Chapter 9 Chemical Quantities in Reactions
9.3
Limiting Reactants
A ceramic brake disc in a sports car withstands temperature of 1400°C.
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Limiting Reactant
A limiting reactant in a chemical reaction is the
substance that • is used up first• stops the reaction• limits the amount of product that can form
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Reacting Amounts
In a table setting, there are 1 fork, 1 knife, and 1 spoon.
How many table settings are possible from 6 forks, 4 spoons, and 7 knives?
What is the limiting item?
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Reacting Amounts
Four table settings are possible.
Initially Used Extra
forks 6 4 2
spoons 4 4 0
knives 7 4 3
The limiting item is the spoon.
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Example of Everyday Limiting Reactant
How many peanut butter sandwiches can be made from 8 slices of bread and 1 jar of peanut butter?
With 8 slices of bread, only 4 sandwiches can be made. The bread is the limiting reactant.
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Example of Everyday Limiting ReactantHow many peanut butter sandwiches can be made from 8 slices of bread and 1 tablespoon of peanut butter?
With 1 tablespoon of peanut butter, only 1 sandwich can be made. The peanut butter is the limiting item.
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Limiting Reactants
When 4.00 mol of H2 is mixed with 2.00 mol of Cl2, how many moles of HCl can form?
H2(g) + Cl2(g) 2HCl (g)
4.00 mol 2.00 mol ??? Mol
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Limiting Reactants Using Moles
Calculate the moles of product from each reactant, H2 and Cl2.
moles of HCl from moles of H2
4.00 mol H2 x 2 mol HCl = 8.00 mol of HCl 1 mol H2 (not possible)
moles of HCl from moles of Cl2
2.00 mol Cl2 x 2 mol HCl = 4.00 mol of HCl 1 mol Cl2 (smaller number)
The limiting reactant is Cl2 because it produces the smaller number of moles of HCl.
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Checking Calculations
Initial
Reactants
H2
4.00 mol
Cl2 2.00 mol
Product
2HCl
0 mol
React/
Form
2.00 mol 2.00 mol +4.00 mol
Left after reaction
2.00 mol
Excess
0 mol
Limiting
4.00 mol
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Calculating Mass of Product from a Limiting Reactant
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Limiting Reactants Using Mass
Calculate the mass of water produced when
8.00 g of H2 and 24.0 g of O2 react?
2H2(g) + O2(g) 2H2O(l)
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Limiting Reactants Using Mass
STEP 1 Use molar mass to convert the grams of each reactant to moles.
Given 8.00 g of H2 and 24.0 g of O2
Need grams of H2O
8.00 g H2 x 1 mol H2 = 3.97 mol of H2
2.016 g H2
24.0 g O2 x 1 mol O2 = 0.750 mol of O2
32.00 g O2
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Limiting Reactants Using Mass (continued)STEP 2 Write mole-mole factors using the
coefficients in the equation.
1 mol of O2 = 2 mol of H2O 1 mol O2 and 2 mol H2O 2 mol H2O 1 mol O2
2 mol of H2 = 2 mol of H2O 2 mol H2 and 2 mol H2O 2 mol H2O 2 mol H2
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Limiting Reactants Using Mass (continued)STEP 3 Calculate moles of product from each
reactant and determine the limiting reactant.
moles of H2O from moles of H2
3.97 mol H2 x 2 mol H2O = 3.97 mol of H2O 2 mol H2 (not possible)
moles of H2O from moles of O2
0.750 mol O2 x 2 mol H2O = 1.50 mol of H2O 1 mol O2 (smaller number)
The limiting reactant is O2 because it produces the smaller moles (1.50 mol) of H2O.
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Limiting Reactants Using Mass (continued)STEP 4 Determine the moles of product or
calculate grams of product using molar mass.
1.50 mol H2O x 18.02 g H2O = 27.0 g of H2O 1 mol H2O
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Limiting
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