Unit 4 Practice: Medium
Chemical Reactions · 20 questions. Try each question before revealing the answer and worked explanation.
Question 1
Write the complete ionic equation for:
AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)
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Answer:
Ag⁺ + NO₃⁻ + Na⁺ + Cl⁻ → AgCl(s) + Na⁺ + NO₃⁻
Explanation: Strong soluble electrolytes are written as separate ions.
AgCl is an insoluble solid, so it remains together.
Question 2
Identify the spectator ions in Question 1.
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Answer: Na⁺ and NO₃⁻
Explanation: These ions appear unchanged on both sides of the ionic equation.
They do not participate in forming AgCl.
Question 3
Write the net ionic equation for a strong acid reacting with a strong base.
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Answer:
H⁺(aq) + OH⁻(aq) → H₂O(l)
Explanation: The essential chemical change in strong acid-strong base neutralization is the combination of H⁺ and OH⁻ to form water.
Question 4
Why is CH₃COOH usually not separated into ions in a net ionic equation?
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Answer: Because CH₃COOH is a weak acid.
Explanation: Weak electrolytes ionize only partially.
In ionic equations, weak acids are usually written mostly as intact molecules.
Question 5
Write the net ionic equation for precipitation of BaSO₄.
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Answer:
Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
Explanation: BaSO₄ is insoluble in water, so the ions combine to form a solid precipitate.
Question 6
For:
2H₂ + O₂ → 2H₂O
How many grams of H₂O form from 3.00 mol O₂?
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Answer: About 108 g H₂O
Work:
Mole ratio:
3.00 mol O₂ × (2 mol H₂O / 1 mol O₂)
= 6.00 mol H₂O
Convert to grams:
6.00 × 18.02 = 108.1 g
Explanation: Always move through the balanced equation in moles before converting to mass.
Question 7
How many moles NaOH are present in 50.0 mL of 0.200 M NaOH?
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Answer: 0.0100 mol
Work:
Convert volume:
50.0 mL = 0.0500 L
Then:
n = MV
n = (0.200)(0.0500)
= 0.0100 mol
Question 8
For:
N₂ + 3H₂ → 2NH₃
If 2.00 mol N₂ and 4.00 mol H₂ are available, which is limiting?
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Answer: H₂
Explanation: 2 mol N₂ would require:
2 × 3 = 6 mol H₂
Only 4 mol H₂ are available.
Therefore H₂ runs out first.
Question 9
How many moles NH₃ form from Question 8?
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Answer: 2.67 mol NH₃
Work:
Use limiting reactant H₂:
4.00 mol H₂ × (2 mol NH₃ / 3 mol H₂)
= 2.67 mol NH₃
Question 10
The theoretical yield is 20.0 g, but the actual yield is 17.0 g. Find percent yield.
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Answer: 85.0%
Work:
(17.0 / 20.0) × 100
= 85.0%
Question 11
Determine the oxidation number of S in SO₄²⁻.
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Answer: +6
Work:
Each O is −2.
Four O:
4(−2)=−8
Total ion charge is −2.
Let S = x:
x − 8 = −2
x = +6
Question 12
Determine the oxidation number of Mn in MnO₄⁻.
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Answer: +7
Work:
Four O atoms:
4(−2)=−8
Total charge:
−1
So:
x − 8 = −1
x = +7
Question 13
In:
Zn + Cu²⁺ → Zn²⁺ + Cu
Which species is oxidized?
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Answer: Zn
Explanation: Zn changes:
0 → +2
It loses 2 electrons, so it is oxidized.
Question 14
What is the oxidizing agent in Question 13?
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Answer: Cu²⁺
Explanation: Cu²⁺ gains electrons and becomes Cu.
The oxidizing agent is the species that causes another species to oxidize while it gets reduced.
Question 15
Write the oxidation half-reaction for Zn.
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Answer:
Zn → Zn²⁺ + 2e⁻
Explanation: Electrons appear on the product side because Zn loses electrons.
Question 16
Identify the conjugate acid-base pairs in:
NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
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Answer:
NH₃ / NH₄⁺
and
H₂O / OH⁻
Explanation: Conjugate acid-base pairs differ by one H⁺.
NH₃ gains H⁺ to become NH₄⁺.
H₂O loses H⁺ to become OH⁻.
Question 17
What is the conjugate base of H₂PO₄⁻?
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Answer: HPO₄²⁻
Explanation: A conjugate base forms when an acid loses H⁺.
Removing H⁺ from H₂PO₄⁻ gives HPO₄²⁻.
Question 18
What is the conjugate acid of NH₃?
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Answer: NH₄⁺
Explanation: Add H⁺:
NH₃ + H⁺ → NH₄⁺
Question 19
A buret reading changes from 2.10 mL to 24.60 mL. How much titrant was delivered?
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Answer: 22.50 mL
Work:
24.60 − 2.10 = 22.50 mL
Explanation: Buret volume delivered is final reading minus initial reading.
Question 20
Why can you not always use M₁V₁ = M₂V₂ in a titration?
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Answer: Because the reacting substances may not have a 1:1 mole ratio.
Explanation: The balanced chemical equation determines the stoichiometric relationship.
If the ratio is not 1:1, moles must be calculated first and then related using the coefficients.