Download Limiting Reactant Practice KEY

Chemistry
HS/Science
Unit: 08 Lesson: 01
Limiting Reactants Practice KEY
1. How many grams of aluminum chloride could be produced from 35.5 g of aluminum and 39.0 g
of chlorine gas?
2Al + 3 Cl2
35.5 g Al X
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2AlCl3
1 mol Al X 3 mol Cl2 X 70.906 g Cl2 = 140. g Cl2
26.982 g Al
2 mol Al
1 mol Cl2
This means 35.5 g Al would react with 140. g Cl2
Only 39.0 g Cl2 were available, so it is the limiting reactant.
Use the mass of Cl2 in the reaction to determine the mass of the AlCl 3 formed.
39.0 g Cl2 =
__
?______ g AlCl3
39.0 g Cl2 X 1 mol Cl2 X
70.906 g Cl2
2 mol AlCl3 X 133.341 g AlCl 3 =
3 mol Cl2
1 mol AlCl2
Answer: 48.9 g AlCl 3
2. Identify the limiting reagent when 10.8 g water reacts with 4.7 g Na to produce NaOH and H 2.
To solve, set one reactant to the other and compare what is calculated to the actual
amount in the reaction to determine the theoretical limiting reactant.
10.8 g H2O = ____?__ g Na
Or
4.7 g Na = ____?__ g H2O
2H2O + 2 Na
2 NaOH + H2
10.8 g H2O X 1 mol H2O x 2 mol Na
18.015g H2O 2 mol H2O
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X 22.990 g Na = 13.8 g Na
1 mol Na
This means that 10.8 g will react completely with 13.8 g Na, but only 4.7 g Na was
available to this reaction. Therefore, Na is the limiting reactant in this problem.
©2012, TESCCC
05/08/13
page 1 of 2
Chemistry
HS/Science
Unit: 08 Lesson: 01
3. If 31.2 g of barium nitrate and 21.9 g of ammonium carbonate are mixed, how many moles of
ammonium nitrate are produced?
Ba(NO3)2 + (NH4)2 CO3
BaCO3 + 2NH4 NO3
After balancing the chemical equation, calculate the limiting reactant and then use the
limiting reactant to determine the mass of the product formed.
31.2 g BaNO3 = __?____ g (NH4)2 CO3
OR
21.9 g (NH4)2CO3 = __?____g BaNO3
31.2 g BaNO3 X 1 mol BaNO3 X 1 mol (NH4)2CO3 X 96.086 g (NH4)2CO3 = 11.5 g (NH4)2CO3
199.332 g BaNO3 1 mol BaNO3
1 mol (NH4)2CO3
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31.2 g BaNO3 would react with 11.5 g (NH4)2 CO3.
Since 21.9 g of (NH4)2 CO3 was available, then it is the excess reactant. Therefore,
BaNO3 is the limiting reactant.
Use the limiting reactant to determine the mass of the product.
31.2 g BaNO3 = _________ g NH4 NO3
31.2 g BaNO3 X 1 mol BaNO3 X 2 mol NH4 NO3
199.332 g BaNO3 1 mol Ba(NO3)2 = 0.313 mol NH4 NO3
©2012, TESCCC
05/08/13
page 2 of 2