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CHEM 121 ACTUAL EXAM QUESTIONS WITH COMPLETE SOLUTIONS VERIFIED 2024/2025 $7.99
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CHEM 121 ACTUAL EXAM QUESTIONS WITH COMPLETE SOLUTIONS VERIFIED 2024/2025
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CHEM 121 ACTUAL EXAM QUESTIONS WITH COMPLETE SOLUTIONS VERIFIED 2024/2025 Consider the hypothetical reaction 3 A + 2 B → 2 C How many mol of B must react with excess A to produce 2 mol of C? 2 (because you must have 2B to make 2 C, 2B is the limiting factor) Consider the following balanc...

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  • June 22, 2024
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  • 2023/2024
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CHEM 121 ACTUAL EXAM QUESTIONS WITH COMPLETE
SOLUTIONS VERIFIED 2024/2025


Consider the hypothetical reaction
3A+2B→2C
How many mol of B must react with excess A to produce 2 mol of C?
2 (because you must have 2B to make 2 C, 2B is the limiting factor)
Consider the following balanced reaction equation:
8 HNO3 + 3 Cu → 2 Cu(NO3)2 + 4 H2O + 2 NO
Which of the following would be correct stoichiometric factors based on this
reaction? Select all that apply.
(8 mol HNO3/4 mol H2O)
(3 mol Cu/2 mol NO)
Consider the following reaction:
4 NH3 + 5 O2 → 4 NO + 6 H2O
If 0.800 mol of O2 reacts via this reaction in excess NH3, how many mol of H2O
will be produced?
(0.800 mol O2 x (6 mol H2O/5 mol O2) = 0.96 mol H2O)
0.96
In a reaction, 2 mol of compound A are needed to react with 3 mol of compound
B. What is the minimum number of mol of compound A required to completely
react with 4.25 mol of compound B? Report your answer to three significant
figures.
(2/3) x 4.25 =
2.83
Consider the following reaction:
SO2 + 2 H2S → 3 S + 2 H2O

, If 7.40 g of SO2 reacts via this reaction in excess H2S, what mass (in g) of S will
be produced?
(7.40 g SO2 x (1 mol SO2/64.07 g SO2) x (3 mol S/1 mol SO2) x (32.07 g S/1 mol S) =
11.1 g S)
11.1
Consider the following reaction:
4 FeS + 7 O2 → 2 Fe2O3 + 4 SO2
If 1.00 mol of FeS and 1.25 mol of O2 are present, how many mol of Fe2O3 can be
produced?
(Find the limiting factor)


(1 mol FeS x (2 mol Fe2O3/2 mol FeS) = 0.5 mol FeO3)


(1.25 mol O2 x (2 mol Fe2O3/7 mol O2) = 0.36 mol Fe2O3)


0.36 (because it is the limiting factor)
Consider the following reaction:
C + 2 H2 → CH4
If 6.25 g C and 6.00 g of H2 are present, what mass of CH4 can be produced
(molar mass CH4 = 16.042 g/mol)
(Find the limiting factor)


(6.25 g C x (1 mol C/12.01 C) x (1 mol CH4/1 mol C) x (16.042 g CH4/1 mol CH4) =
8.35 g CH4)


(6.00 g H2 x (1 mol H2/2.016 g H2) x (1 mol CH4/2 mol H2) x (16.042 g CH4/ 1 mol
CH4) = 23.9 CH4)


8.35 (because it is the limiting factor)
Consider the following balanced reaction equation:
N2 + O2 → 2 NO

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