Name: _____________
Period: _____
Stoichiometry Basics and Practice - KEY
I will not be in class on Tuesday, 02/17/2013, while you are taking your exam (Day 2). Email me at mdo@hondoisd.org
or mdo@hondoisd.net
with questions/concerns.
Be able to calculate/obtain the following:

Molar mass (a link between GRAMS and MOLES): calculated via Period Table, specifically atomic mass, taking into consideration the number of atoms (subscripts) and types of atoms (elements) in the compound; units are in GRAMS PER MOLE (GRAMS/MOLE)

Molar ratio (a link between MOLES and MOLES): obtained from the coefficients (large numbers) in front of elements and compounds in a BALANCED equation
Balance the following equations:
_1_ N
2
+ _3_ F
2
 _2__ NF
3
_1_ HBr + _1_ KHCO
3
 _1_ H
2
O + _1_ KBr + _1_ CO
2
From the following, give me the requested mole ratios/appropriate amounts:
2 C
6
H
10
+ 17 O
2
 12 CO
2
+ 10 H
2
O
Mole ratio of oxygen to carbon dioxide: 17 moles of oxygen to 12 moles of carbon dioxide
Mole ratio of oxygen to water: 17 moles of oxygen to 10 moles of water
Mole ratio of C
6
H
10 to O
2:
2 moles of C
6
H
10 to 17 moles of oxygen
If I have 2 moles of C
6
H
10
, how many moles of oxygen (O
2
) I need to completely react with this amount? I will need
17 moles of oxygen to completely react with the 2 moles of C
6
H
10 that I have. I obtained this from the mole ratio of 2 moles of C
6
H
10 to 17 moles of oxygen.
If I have 4 moles of C
6
H
10
, how many moles of water will be produced? This reaction will product 20 moles of water. I obtained this from the mole ratio of 2 moles of C
6
H
10 to 10 moles of water.
Using the equation 2 C
6
H
10
+ 17 O
2
 12 CO
2
+ 10 H
2
O, answer the following questions:
If I do this reaction with 35 grams of C
6
H
10
and 45 grams of oxygen, how many grams of carbon dioxide will be formed? Given two amounts of reactants, you should think: LIMITING REAGENT (Recall: spinach/omelet analogy)
35 grams C
6
H
10
1 mole C
6
H
10
12 moles CO
2
44.009 g CO
2
= 112.506 g CO
2
82.146 grams C
6
H
10
2 moles C
6
H
10
1 mole CO
2
45 grams O
2
1 mole O
2
31.998 grams O
2
12 moles CO
17 moles O
2
2
44.009 g CO
1 mole CO
2
2
= 43.6881 g CO
2
Based on my two stoichiometric setups, I can produce 43.6881 g CO
2
. I pick the lesser of the two values because one of the reactants (to be discussed below) limits me/holds me back in how much I can make.

What is the limiting reagent in the above reaction? Oxygen is the limiting reagent or limiting reactant. With these two given amounts, it can be said also that C
6
H
10 is in excess.
If 35 grams of carbon dioxide are actually formed from the reaction in problem 5, what is the percent yield of this reaction? % yield = (actual yield) / (theoretical yield) x 100
Actual yield = 35 grams (given in the problem)
Theoretical yield = 43.6881 grams (we can think of theoretical yield as “calculated yield” – we calculated it above, taking into consideration the limiting reagent)
% yield = 35 grams/43.6881 grams x 100 = 80.11% yield
Using the following equation:
2 NaOH + H
2
SO
4
 2 H
2
O + Na
2
SO
4
How many grams of sodium sulfate will be formed if you start with 200 grams of sodium hydroxide and you have an excess of sulfuric acid? When you see the word “excess”, you should think “This means that the other compound is the limiting reagent.”
200 g NaOH 1 mole NaOH
39.997 g NaOH
1 mole Na
2
SO4
2 moles NaOH
142.042 grams Na
1 mole Na
2
SO
4
2
SO
4
= 355.132 grams Na
2
SO
4
Using the following equation:
Pb(SO
4
)
2
+ 4 LiNO
3
 Pb(NO
3
)
4
+ 2 Li
2
SO
4
How many grams of lithium nitrate will be needed to make 250 grams of lithium sulfate, assuming that you have an adequate amount of lead (IV) sulfate to do the reaction?
250 grams Li
2
SO
4
1 mole Li
2
SO
4
4 moles LiNO
3
68.945 grams LiNO
3
= 313.546 grams LiNO
3
109.944 grams Li
2
SO
4
2 moles Li
2
SO
4
1 mole LiNO
3
Determine the number of moles in 246.0 g of potassium permanganate [KMnO
4
].
246 g KMnO
4
1 mole KMnO
4
158.032 grams KMnO
4
= 1.557 moles KMnO
4
Silver nitrate reacts with sodium chloride left from the perspiration on the skin and forms solid silver chloride and sodium nitrate. How many grams of sodium nitrate are formed when 0.230 grams of sodium chloride react?
AgNO
3
+ NaCl  AgCl + NaNO
3
0.23 g NaCl 1 mole NaCl
58.443 grams NaCl
1 mole NaNO
1 mole NaCl
3
84.994 grams NaNO
1 mole NaNO
3
3
= 0.334 g NaNO
3

274.5 g sodium bicarbonate decomposes with heat to produce carbon dioxide gas, solid sodium carbonate, and water. What is the percent yield if you captured and condensed 1.1 g of water as products of the reaction?
2 NaHCO
3
 CO
2
+ Na
2
CO
3
+ H
2
O
% yield = (actual yield) / (theoretical yield) x 100
Actual yield = 1.1 grams water
Theoretical yield…we must calculate this! There is fortunately no limiting reagent really here since there is only one reactant in our equation.
Sodium bicarbonate is NaHCO
3
274.5 grams sodium bicarbonate
1 mole sodium bicarbonate 1 mole water
18.015 grams water
84.006 grams sodium bicarbonate
WE ARE NOT DONE YET!
29.4331 grams is our theoretical yield.
2 moles sodium bicarbonate
1 mole water
% yield = 1.1 grams water/29.4331 grams water x 100 = 3.737 % yield
=
29.4331 grams water