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General Chemistry, CHEM 101
Exam 1, 02/25/2021 @08:00 AM
Exam duration is 40 minutes
The exam needs to be uploaded to Canvas in one single file before 08:50 am
Honor code statement: I certify that the work is my own, and I did not open the textbook, notes,
or any other source during the time of this exam. Signature _______________
Problem 1. (30 points)
PART 1. Write the FULL IONIC equation for each of the following reactions? (20 pnts)
A)
K2SO4(aq) + 2 AgNO3(aq) → 2 KNO3(aq) + Ag2SO4(s)
B) Na2CO3(aq) + 2 HCl(aq) → 2 NaCl(aq) + CO2(g) + H2O(l)
PART 2. Write the NET IONIC equations of the previous two reactions? (10 pnts)
A)
B)
Problem 2. (30 points)
PART 1. Consider the following octane (C8H18) combustion reaction:
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
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If the reaction consumes 60 moles of O2 to burn C8H18, how many moles of CO2 are formed? (15
pnts)
16 𝑚𝑜𝑙 𝐶𝑂2 960
=
𝑚𝑜𝑙 𝑜𝑓 𝐶𝑂2 = 38.4
25 𝑚𝑜𝑙 𝑂2
25
= 38 𝑜𝑟 39 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶𝑂2
60 𝑚𝑜𝑙 𝑜𝑓 𝑂2 ×
either 38.4, 38, or 39 moles of CO2 you will get the credit
PART 2. Consider the following reaction
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
If we have 10 molecules of CH4 and 16 molecules of O2, which reactant is the limiting reactant?
(15 pnts)
𝐶𝐻4 :
1𝐶𝑂2
10 𝐶𝐻4 ×
= 10 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝑜𝑓 𝐶𝑂2
1𝐶𝐻4
1𝐶𝑂2
= 8 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝐶𝑂2
𝟐𝑂2
Please notice that according to the reaction equation, 2 moles of
O2 are required to make 1 mole of CO2.
We have enough CH4 to make 10 molecules of CO2
We have enough O2 to make 8 molecules of CO2, Therefore, O2
is the limiting reactant (you can find the limiting reactant using
the H2O instead of using CO2, you will be correct)
𝑂2 :
16 𝑂2 ×
Problem 3. (40 points, each question = 4 pnts)
Question #
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Answer
C
D
B
B
B
A
E
C
B
D
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