CHEM 121 ACTUAL EXAM QUESTIONS WITH COMPLETE SOLUTIONS VERIFIED 2024/2025
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Module
CHEM 121
Institution
CHEM 121
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...
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)
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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