,Portage Learning CHEM 103 General
Chemistry I Module 4 Exam
Question 1
When the chemical equation \(\underline{\text{\ }}\text{Al}+\underline{\text{\
}}\text{O}_{2}\rightarrow \underline{\text{\ }}\text{Al}_{2}\text{O}_{3}\) is correctly
balanced using the lowest whole number coefficients, what is the coefficient for
\(\text{O}_{2}\)?
A) 1
B) 2
C) 3
D) 4
Answer: C
Rationale: To balance the oxygen atoms, find the least common multiple of 2 and 3,
which is 6. Placing a coefficient of 3 in front of \(\text{O}_{2}\) gives 6 oxygen atoms on
the reactant side, and placing a 2 in front of \(\text{Al}_2\text{O}_3\) gives 6 oxygen
atoms on the product side. This creates 4 aluminum atoms on the product side,
requiring a coefficient of 4 in front of \(\text{Al}\). The fully balanced equation is
\(4\text{Al}+3\text{O}_{2}\rightarrow 2\text{Al}_{2}\text{O}_{3}\).
Question 2
What is the sum of all coefficients when the following reaction is balanced with the
lowest whole number coefficients: \(\underline{\text{\
}}\text{C}_{3}\text{H}_{8}+\underline{\text{\ }}\text{O}_{2}\rightarrow \underline{\text{\
}}\text{CO}_{2}+\underline{\text{\ }}\text{H}_{2}\text{O}\)?
A) 10
B) 13
C) 11
D) 12
Answer: B
Rationale: First, balance the carbons by placing a 3 in front of \(\text{CO}_{2}\). Next,
balance the hydrogens by placing a 4 in front of \(\text{H}_2\text{O}\). Count the
oxygens on the product side: \((3 \times 2) + (4 \times 1) = 10\) oxygen atoms. To get 10
oxygens on the reactant side, place a 5 in front of \(\text{O}_{2}\). The balanced
, equation is \(1\text{C}_{3}\text{H}_{8}+5\text{O}_{2}\rightarrow
3\text{CO}_{2}+4\text{H}_{2}\text{O}\). Sum of coefficients \(= 1 + 5 + 3 + 4 = 13\).
Question 3
How many moles of \(\text{CO}_{2}\) are produced when 2.50 moles of
\(\text{C}_3\text{H}_8\) are burned completely in excess oxygen according to the
balanced equation: \(\text{C}_{3}\text{H}_{8}+5\text{O}_{2}\rightarrow
3\text{CO}_{2}+4\text{H}_{2}\text{O}\)?
A) 2.50 moles
B) 7.50 moles
C) 1.20 moles
D) 5.00 moles
Answer: B
Rationale: Use the molar ratio between \(\text{C}_3\text{H}_8\) and \(\text{CO}_{2}\)
from the balanced equation, which is 3 moles of \(\text{CO}_{2}\) for every 1 mole of
\(\text{C}_3\text{H}_8\). Calculation: \(2.50\text{ moles C}_3\text{H}_8 \times (3\text{
moles CO}_\text{ mole C}_3\text{H}_8) = 7.50\text{ moles CO}_2\).
Question 4
Given the reaction \(\text{N}_{2}+3\text{H}_{2}\rightarrow 2\text{NH}_{3}\), how many
grams of \(\text{NH}_{3}\) (Molar Mass \(= 17.03\text{ g/mol}\)) can be produced from
0.600 moles of \(\text{H}_{2}\) acting with excess \(\text{N}_{2}\)?
A) 6.81 g
B) 20.4 g
C) 10.2 g
D) 15.3 g
Answer: A
Rationale: First, convert moles of \(\text{H}_{2}\) to moles of \(\text{NH}_{3}\) using the
mole ratio: \(0.600\text{ moles H}_2 \times (2\text{ moles NH}_\text{ moles H}_2) =
0.400\text{ moles NH}_3\). Next, convert moles of \(\text{NH}_{3}\) to grams:
\(0.400\text{ moles} \times 17.03\text{ g/mol} = 6.81\text{ g}\).
Question 5
If a reaction has a theoretical yield of 45.2 grams, but a student successfully isolates
Chemistry I Module 4 Exam
Question 1
When the chemical equation \(\underline{\text{\ }}\text{Al}+\underline{\text{\
}}\text{O}_{2}\rightarrow \underline{\text{\ }}\text{Al}_{2}\text{O}_{3}\) is correctly
balanced using the lowest whole number coefficients, what is the coefficient for
\(\text{O}_{2}\)?
A) 1
B) 2
C) 3
D) 4
Answer: C
Rationale: To balance the oxygen atoms, find the least common multiple of 2 and 3,
which is 6. Placing a coefficient of 3 in front of \(\text{O}_{2}\) gives 6 oxygen atoms on
the reactant side, and placing a 2 in front of \(\text{Al}_2\text{O}_3\) gives 6 oxygen
atoms on the product side. This creates 4 aluminum atoms on the product side,
requiring a coefficient of 4 in front of \(\text{Al}\). The fully balanced equation is
\(4\text{Al}+3\text{O}_{2}\rightarrow 2\text{Al}_{2}\text{O}_{3}\).
Question 2
What is the sum of all coefficients when the following reaction is balanced with the
lowest whole number coefficients: \(\underline{\text{\
}}\text{C}_{3}\text{H}_{8}+\underline{\text{\ }}\text{O}_{2}\rightarrow \underline{\text{\
}}\text{CO}_{2}+\underline{\text{\ }}\text{H}_{2}\text{O}\)?
A) 10
B) 13
C) 11
D) 12
Answer: B
Rationale: First, balance the carbons by placing a 3 in front of \(\text{CO}_{2}\). Next,
balance the hydrogens by placing a 4 in front of \(\text{H}_2\text{O}\). Count the
oxygens on the product side: \((3 \times 2) + (4 \times 1) = 10\) oxygen atoms. To get 10
oxygens on the reactant side, place a 5 in front of \(\text{O}_{2}\). The balanced
, equation is \(1\text{C}_{3}\text{H}_{8}+5\text{O}_{2}\rightarrow
3\text{CO}_{2}+4\text{H}_{2}\text{O}\). Sum of coefficients \(= 1 + 5 + 3 + 4 = 13\).
Question 3
How many moles of \(\text{CO}_{2}\) are produced when 2.50 moles of
\(\text{C}_3\text{H}_8\) are burned completely in excess oxygen according to the
balanced equation: \(\text{C}_{3}\text{H}_{8}+5\text{O}_{2}\rightarrow
3\text{CO}_{2}+4\text{H}_{2}\text{O}\)?
A) 2.50 moles
B) 7.50 moles
C) 1.20 moles
D) 5.00 moles
Answer: B
Rationale: Use the molar ratio between \(\text{C}_3\text{H}_8\) and \(\text{CO}_{2}\)
from the balanced equation, which is 3 moles of \(\text{CO}_{2}\) for every 1 mole of
\(\text{C}_3\text{H}_8\). Calculation: \(2.50\text{ moles C}_3\text{H}_8 \times (3\text{
moles CO}_\text{ mole C}_3\text{H}_8) = 7.50\text{ moles CO}_2\).
Question 4
Given the reaction \(\text{N}_{2}+3\text{H}_{2}\rightarrow 2\text{NH}_{3}\), how many
grams of \(\text{NH}_{3}\) (Molar Mass \(= 17.03\text{ g/mol}\)) can be produced from
0.600 moles of \(\text{H}_{2}\) acting with excess \(\text{N}_{2}\)?
A) 6.81 g
B) 20.4 g
C) 10.2 g
D) 15.3 g
Answer: A
Rationale: First, convert moles of \(\text{H}_{2}\) to moles of \(\text{NH}_{3}\) using the
mole ratio: \(0.600\text{ moles H}_2 \times (2\text{ moles NH}_\text{ moles H}_2) =
0.400\text{ moles NH}_3\). Next, convert moles of \(\text{NH}_{3}\) to grams:
\(0.400\text{ moles} \times 17.03\text{ g/mol} = 6.81\text{ g}\).
Question 5
If a reaction has a theoretical yield of 45.2 grams, but a student successfully isolates
