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ChemQuest 33
Name: ____________________________
Date: _______________
Hour: _____
Information: Limiting Reactant
Again consider the combustion of propane: C3H8 + 5 O2 3 CO2 + 4 H2O. If you had 10 moles
of propane to burn, you would need 50 moles of oxygen according to the ratio in the balanced
equation. If you only had 20 moles of oxygen you could not combust all 10 moles of propane. The
reaction has been limited by the amount of oxygen you have—you don’t have enough oxygen to
burn all of the propane. In this case, oxygen is called the “limiting reactant” because it limits how
much propane can react. Notice that the limiting reactant isn’t always the substance that is present in
the fewest number of moles. In this example, propane (C3H8) is the “excess reactant” because after
the reaction there will be some of it left over. It is important to remember that everything in a
chemical equation is related by mole ratios. If you only know the mass (grams) of the substances,
you need to convert to moles.
Critical Thinking Questions
8. a) In the above discussion, it was evident that 20 moles of oxygen was not sufficient to combust
10 moles of propane. How many moles of the propane can be combusted with 20 moles of
oxygen?
4 moles. For every 5 moles of oxygen, 1 mole of propane can be combusted, so the number of
moles of propane that can be combusted with 20 moles of oxygen is 20/5 = 4 moles
b) How many moles of carbon dioxide will be produced? (Base the answer to this question on the
number of moles of propane that actually get combusted—which is your answer to part a.)
12 moles. For every mole of propane that combusts 3 moles of CO2 are produced, so the number
of moles of CO2 that can be produced when 4 moles of propane combusts = 4(3) = 12.
c) Verify that if 12.5 moles of propane and 63.2 moles of oxygen were present, then propane is
the “limiting reactant” and oxygen is the excess reactant.
For each mole of C3H8 five moles of O2 are required, so for 12.5 moles of C3H8, the number
of moles of O2 needed are (12.5)(5) = 62.5 moles. Since we have more than 62.5 moles
(according to the question we have 63.2 moles), then we have excess O2 and therefore C3H8 is
the limiting reactant.
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9. Consider the following chemical reaction: 3 MgCl2 + 2 Na3PO4 6 NaCl + Mg3(PO4)2.
Assume that 0.75 mol of MgCl2 and 0.65 mol of Na3PO4 are placed in a reaction vessel.
a) Verify that Na3PO4 is the excess reactant and MgCl2 is the limiting reactant.
Using the 3:2 ratio, we find that for 0.75 mol of MgCl2, we need (0.75)(2/3) = 0.5 moles of
Na3PO4. We have more than that and therefore Na3PO4 is the excess reactant and MgCl2 the
limiting.
b) How many moles of the excess reactant are left over after the reaction stops?
We have 0.65 mol and we use up 0.5 mol (calculated in part a), so the number of moles left
over is 0.65 – 0.5 = 0.15 mol
c) How many moles of NaCl will be produced in this reaction? (Remember—you must base this
answer on how many moles of the limiting reactant that reacted.)
1.5 mol. 0.75 mol of MgCl2 is the limiting reactant, so we use the 6:3 ratio of NaCl to MgCl2
and multiply (0.75)(6/3) = 1.5 mol.
10. Consider the double replacement reaction between calcium sulfate (CaSO4) and sodium iodide
(NaI). If 34.7 g of calcium sulfate and 58.3 g of sodium iodide are placed in a reaction vessel,
how many grams of each product are produced? (Hint: Do this problem in the steps outlined
below.)
a) Write the balanced chemical equation for the reaction.
CaSO4 + 2 NaI CaI2 + Na2SO4
b) Find the limiting reactant. First, convert 34.7g and 58.3g from grams to moles using the
molar masses from the periodic table. Next, compare the number of moles of each reactant.
Ask yourself: Do I have enough NaI to use up all of the CaSO4? Do I have enough CaSO4 to
use up all of the NaI? Whichever one will get used up is the limiting reactant.
(34.7g) ÷ (136.2g/mol) = 0.255 mol CaSO4; (58.3g) ÷ (149.9g/mol) = 0.389 mol NaI
To use up all of the NaI, I need (0.389)(½) = 0.1945 mol CaSO4 I have more than this
much, so NaI is limiting reactant.
c) Use the number of moles of the limiting reactant to calculate the number of moles of each
product produced using the coefficients from the balanced chemical equation in part a.
mol of CaI2 = mol of Na2SO4 because the coefficients are both “1” in the balanced equation.
Using the mol of limiting reactant and the mole ratio of 1:2, you can calculate the mol of CaI2
and Na2SO4. (0.389)(½) = 0.1945 mol.
d) In part c you found the moles of each product produced. Now convert moles to grams using
the molar mass from the periodic table. You have now answered the question.
g of CaI2 = (0.1945 mol)(293.9g/mol) = 57.2g
g of Na2SO4 = (0.1945 mol)(142.1g/mol) = 27.6g
Copyright 2002-2004 by Jason Neil. All rights reserved.
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11. If 181.1g of Al(NO3)3 react with 102.1g of CaO in a double replacement reaction, how many
grams of each product will be produced? (Note: this is just like the last question, but parts a-d are
not spelled out for you.)
First, write the balanced equation:
2 Al(NO3)3 + 3 CaO 3 Ca(NO3)2 + Al2O3
Next, find the limiting reactant:
(181.1g) ÷ (213 g/mol) = 0.850 mol Al(NO3)3
(102.1g) ÷ (56.1g/mol) = 1.82 mol CaO
To use up all 0.850 mol of Al(NO3)3, I need (0.850)(3/2) = 1.275 mol CaO. Since you have
more than this amount, CaO is present in excess and Al(NO3)3 is the limiting reactant.
Use the moles of limiting reactant to calculate the moles of each product produced:
mol Ca(NO3)2 = (0.850)(3/2) = 1.275 mol
mol Al2O3 = (0.850)(½) = 0.425 mol
Convert the mol of each product to grams of each product:
Grams Ca(NO3)2 = (1.275 mol)(164.1g/mol) = 209.2 g
Grams Al2O3 = (0.425mol)(102g/mol) = 43.4 g
Skill Practice 32
Name: ______________________________
Date: _______________
Hour: _____
1. Consider the reaction in which 410 g of Ca(NO3)2 react with just the right amount of lithium
metal in a single replacement reaction.
a) How many grams of lithium are required?
35.0 g
b) How many grams of each product can be produced?
345 g LiNO3
100g Ca
2. When 250 g of Na2SO4 reacts with plenty of Ca3P2 according to the following balanced equation,
how many grams of Na3P will be produced?
2 Na3P + 3 CaSO4
3 Na2SO4 + Ca3P2
130g Na3P
265g CaSO4
3. Consider the following combustion reaction: 2 C4H10 + 13 O2
8 CO2 + 10 H2O. If 75.3 g of
C4H10 react with plenty of O2, what mass of CO2 and H2O can be produced?
229g CO2
117g H2O
4. Consider the following balanced equation: 3Ca(NO3)2 + 2AlCl3 2Al(NO3)3 + 3CaCl2. If
210.5g of calcium nitrate react, what is the mass of each product that can be produced?
409.8g Al(NO3)3
142g CaCl2
5. When 53.6 g of calcium carbonate react with plenty of aluminum fluoride, how many grams of
each product can be produced?
41.8g of each product will be produced
© 2004 by Jason Neil. All rights reserved.