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Chemistry Unit 8 Worksheet 3: Adjusting to Reality - Limiting Reactant

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Chemistry Unit 8
Worksheet 3: Adjusting to Reality - Limiting Reactant
1. Write the balanced equation for the reaction between hydrogen and oxygen.
Balanced Equation: 2 H2(g) + 1 O2(g)
2 H2O(g)
Suppose that 4 molecules of hydrogen gas and 4 molecules of oxygen gas react to
form water.
Make a drawing that represents the reaction container before and after the
reaction.
4 molecules How many molecules of water can be produced?
oxygen
Which reactant is in excess? Why? There are leftover molecules
2 molecules How many molecules of excess reactant are there?
Construct a Before-Change-After Table for this reactant mixture:
In this table the numbers refer to molecules rather than moles.
Equation:
2 H2(g) + 1 O2(g)
2 H2O(g)
Before
4
4
0
Change
-4
-2
+4
After
0
2.
4
According to the table you just made,
4 molecules How many molecules of water can be produced?
oxygen
Which reactant is in excess? Why? 2 molecules are left-over “after” the
reaction
2 molecules How many molecules of excess reactant are there?
Based on your two methods of analysis above, what determines how much product can
be made from a particular reactant mix?
The determining factor is which of the reactants runs out first. This determines how much
product can be made, even if another reactant is in excess.
©Modeling Instruction – AMTA 2013
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U8 ws 3 v2.0
2. Write the equation for the formation of ammonia from nitrogen gas and hydrogen
gas.
Balanced Equation: 1 N2(g) + 3 H2(g)
2 NH3(g)
Given 6 molecules of nitrogen and 12 molecules of hydrogen, make a drawing that
represents the reaction container before and after the reaction.
8 molecules How many molecules of ammonia can be produced?
nitrogen
Which reactant is in excess? Why? Some is leftover after the reaction.
2 molecules How many molecules of excess reactant are there?
Construct a Before-Change-After Table for this reactant mixture:
In this table the numbers refer to molecules rather than moles.
Equation:
Before
1 N2(g)
6
+
3 H2(g)
2 NH3(g)
12
0
Change
-4
-12
+8
After
2
0
8
According to the table you just made,
8 molecules How many molecules of ammonia can be produced?
nitrogen
Which reactant is in excess? Why? Some is leftover after the rxn.
2 molecules How many molecules of excess reactant are there?
Describe what you must look for in a particular reactant mixture to decide which
reactant will be in excess (have some left over after the reaction):
To determine which reactant will be in excess you have to see which one can be
consumed completely with some of the other left over. You do this by using the mole
ratios to figure out how much of each you would need, considering each as the limiting
reactant. Only one reactant will work out to have excess after the reaction.
©Modeling Instruction – AMTA 2013
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U8 ws 3 v2.0
3. When 0.50 mole of aluminum reacts with 0.72 mole of iodine to form aluminum
iodide, how many moles of the excess reactant will remain? 0.02 mol Al
How many moles of aluminum iodide will be formed? 0.48 mol AlI3
2 Al(s)
+
3 I2(s)
Equation:
Before
0.50
0.72
Change
-0.48
-0.72
After
0.02
2 AlI3(s)
0.50 mol Al x 3 mol I2 = 0.75 mol I2
2 mol Al
0
+0.48
0
0.72 mol I2 x 2 mol Al = 0.48 mol Al
3 mol I2
0.48
4. When sodium hydroxide reacts with sulfuric acid (H2SO4), water and sodium
sulfate are the products. Calculate the mass of sodium sulfate produced when
15.5 g of sodium hydroxide are reacted with 46.7 g of sulfuric acid. [Hint: which unit
is used in all stoichiometry reasoning?]
Equation:
2 NaOH(aq)
+
1 H2SO4(aq)
1 Na2SO4(aq)
+
Before
0.388
0.476
Change
-0.388
-0.194
+0.194
+0.388
0.282
0.194
0.3875
After
0
15.5 g NaOH
1mol
40.0g
0.388mol NaOH
0.388mol
1mol Na2SO4
2 mol NaOH
0
2 H2O(l)
46.7 g H2SO4
0.194 mol Na2SO4
1mol
98.1g
142.1g
1mol
0
0.476mol NaOH is limiting
27.6 g
5. A 22.4 g sample of oxygen gas is placed in a sealed container with 2.50 g of
hydrogen gas. The mixture is sparked, producing water vapor. Calculate the mass
of water formed. Calculate the number of moles of the excess reactant remaining.
Equation:
1 O2(g)
Before
0.7
Change
-0.625
After
1.25 mol H 2
.075
2mol H 2 O
2mol H 2
+
2 H2(g)
2 H2O(g)
1.25
0
22.4 g O2 x 1 mol O2 = 0.7 mol O2
32.0 g O2
-1.25
+1.25
0
1.25
1.25mol H 2 O
18 .0 g
1mol
2.5 g H2 x 1 mol H2 = 1.25 mol H2
2.0 g H2
22 .5g H2O are produced
1.25 mol H20 produced x 1 mol O2 = .625 mol O2 consumed
2 mol H20
0.7 mole O2
- 0.625
0.075 moles of O2 remain.
©Modeling Instruction – AMTA 2013
3
U8 ws 3 v2.0
6. Neuroscientists believe that the only chemical in chocolate that may have a feel-good effect
on the human brain is phenylethylamine (PEA). Although the PEA in chocolate occurs
naturally, PEA can be made in the laboratory by the following
reaction:
CH5NO2
+
ammonium formate
C8H8O
пѓ C8H11N
acetophenone
+
CO2
+ H2O
phenylethylamine
(PEA)
phenylethylamine
How much PEA can be made from 75.0g of ammonium formate and
125g of acetophenone? What mass of the excess reactant remains?
Equation:
CH5NO2
+
C8H8O пѓ C8H11N
Before
1.19
1.04
0
Change
-1.04
1.04
+1.04
After
0.15
0
1.04
+
CO2
0
+ H2O
0
75.0 g CH5NO 2
1mol
1.19mol
63.0 g
1.04 mol C8H 8O
1mol C8H11N
121g
1.04 mol C8H11N
126g C8H11N produced
1mol C8H 8O
1mol
0.15mol CH5NO 2
4. 27.5 g
63.0 g
1mol
125g C8H 8O
1mol
1.04 mol
120g
9.45g ammonium formate remain
5. 16.4 g, 0.34 moles xs
©Modeling Instruction – AMTA 2013
6. 126 g, 9.45g xs
4
U8 ws 3 v2.0
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