Worksheet covering quantitative chemistry topics including balancing chemical equations, identifying reaction types, writing ionic equations, and calculating mass of products from combustion reactions.

Let’s go through each question one by one.

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Question 1:

We are asked to classify reaction types, balance equations, and write simple equations including states.

Let’s balance each equation and identify the type:

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a) C₄H₁₀(g) + O₂(g) → CO₂(g) + H₂O(g)

This is a combustion reaction (hydrocarbon + oxygen → CO₂ + H₂O).

Balancing:

C₄H₁₀ has 4 C → need 4 CO₂
C₄H₁₀ has 10 H → need 5 H₂O (since each has 2 H)

Now count O on right:
4 CO₂ → 8 O
5 H₂O → 5 O
Total = 13 O atoms → so we need 13/2 O₂ on left.

Multiply entire equation by 2 to eliminate fraction:

→ 2 C₄H₁₀(g) + 13 O₂(g) → 8 CO₂(g) + 10 H₂O(g)

Type: Combustion

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b) AgNO₃(aq) + CuSO₄(aq) → Ag₂SO₄(s) + Cu(NO₃)₂(aq)

This is a double displacement (precipitation) reaction — Ag₂SO₄ is solid precipitate.

Balance:

Left: Ag=1, NO₃=1, Cu=1, SO₄=1
Right: Ag=2, SO₄=1, Cu=1, NO₃=2

So we need 2 AgNO₃ on left to get 2 Ag and 2 NO₃.

→ 2 AgNO₃(aq) + CuSO₄(aq) → Ag₂SO₄(s) + Cu(NO₃)₂(aq)

Type: Double displacement / Precipitation

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c) Cl₂(g) + Na₂S(aq) → NaCl(aq) + S₈(s)

This is a single displacement reaction — chlorine displaces sulfur from sodium sulfide.

But note: product is S₈ (elemental sulfur), not just S.

Balance:

Left: Cl₂, Na₂S
Right: NaCl, S₈

To make S₈, we need 8 S atoms → so 8 Na₂S

That gives 16 Na → so 16 NaCl

16 NaCl needs 8 Cl₂ (since each Cl₂ gives 2 Cl)

→ 8 Cl₂(g) + 8 Na₂S(aq) → 16 NaCl(aq) + S₈(s)

Wait — let’s check:

Left: Cl = 16, Na = 16, S = 8
Right: Na = 16, Cl = 16, S = 8 → balanced!

Type: Single displacement

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d) Ag(s) + O₂(g) + H₂(g) → AgOH(s)

This looks like a synthesis reaction forming silver hydroxide.

Balance:

AgOH has 1 Ag, 1 O, 1 H

O₂ has 2 O → so we need 2 AgOH to use 1 O₂

2 AgOH → 2 Ag, 2 O, 2 H

H₂ provides 2 H → good

So:

→ 2 Ag(s) + ½ O₂(g) + H₂(g) → 2 AgOH(s)

But usually we avoid fractions → multiply by 2:

→ 4 Ag(s) + O₂(g) + 2 H₂(g) → 4 AgOH(s)

Check:
Left: Ag=4, O=2, H=4
Right: Ag=4, O=4? Wait — 4 AgOH has 4 O? No! AgOH has one O per formula → 4 AgOH has 4 O? But we only have O₂ → 2 O atoms.

Mistake here.

AgOH: Ag, O, H → one oxygen per molecule.

If we want 2 AgOH → 2 O atoms → that comes from 1 O₂ → good.

Hydrogen: 2 AgOH → 2 H → from 1 H₂ → good.

Silver: 2 Ag → from 2 Ag(s)

So correct balanced equation:

→ 2 Ag(s) + ½ O₂(g) + H₂(g) → 2 AgOH(s)

Or without fractions:

→ 4 Ag(s) + O₂(g) + 2 H₂(g) → 4 AgOH(s) — but now oxygen: left = 2 O, right = 4 O → NOT balanced.

Ah! Error.

Each AgOH has ONE oxygen atom.

So for 2 AgOH → 2 oxygen atoms → requires 1 O₂ molecule → correct.

But 2 AgOH also has 2 hydrogen atoms → requires 1 H₂ molecule → correct.

And 2 Ag atoms → from 2 Ag(s).

So:

2 Ag(s) + ½ O₂(g) + H₂(g) → 2 AgOH(s) is correct.

To avoid fraction: multiply by 2:

4 Ag(s) + O₂(g) + 2 H₂(g) → 4 AgOH(s) — now check oxygen: left = 2 O, right = 4 O → still wrong.

Wait — no: 4 AgOH has 4 oxygen atoms? Yes! Each AgOH has one O → 4 AgOH = 4 O atoms.

But O₂ provides only 2 O atoms → contradiction.

I see the mistake.

Actually, AgOH is Ag-O-H → so one oxygen per formula unit.

So if we produce 2 AgOH, we need 2 oxygen atoms → which is exactly what 1 O₂ gives.

Similarly, 2 hydrogen atoms from 1 H₂.

And 2 silver atoms from 2 Ag.

So:

2 Ag(s) + ½ O₂(g) + H₂(g) → 2 AgOH(s) — this is chemically correct.

But since we usually write whole numbers, we can leave it as is or accept fractional coefficients temporarily.

Alternatively, perhaps the intended product is different? But based on given, we’ll go with:

→ 2 Ag(s) + ½ O₂(g) + H₂(g) → 2 AgOH(s)

Type: Synthesis

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Now, the next part:

Lead(II) nitrate solution and potassium iodide solution react to produce a precipitate (and another product). Write (including states), balance, and state reaction type.

Lead(II) nitrate: Pb(NO₃)₂(aq)
Potassium iodide: KI(aq)

They swap ions: Pb²⁺ pairs with I⁻ → PbI₂ (which is insoluble — yellow precipitate)
K⁺ pairs with NO₃⁻ → KNO₃ (soluble)

Equation:

Pb(NO₃)₂(aq) + KI(aq) → PbI₂(s) + KNO₃(aq)

Balance:

Pb: 1=1
NO₃: 2 on left → need 2 KNO₃ on right → so 2 KI on left

Then K: 2 on left → 2 on right
I: 2 on left → 2 in PbI₂

→ Pb(NO₃)₂(aq) + 2 KI(aq) → PbI₂(s) + 2 KNO₃(aq)

Type: Double displacement / Precipitation

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Question 2:

Write non-ionic, total ionic, and net ionic equations for barium chlorate + sodium phosphate.

First, formulas:

Barium chlorate: Ba(ClO₃)₂
Sodium phosphate: Na₃PO₄

They react to form barium phosphate (insoluble) and sodium chlorate (soluble).

Barium phosphate: Ba₃(PO₄)₂ (solid precipitate)
Sodium chlorate: NaClO₃ (aqueous)

Non-ionic (molecular) equation:

3 Ba(ClO₃)₂(aq) + 2 Na₃PO₄(aq) → Ba₃(PO₄)₂(s) + 6 NaClO₃(aq)

Check balance:

Ba: 3=3
ClO₃: 6=6
Na: 6=6
PO₄: 2=2 → good.

Total ionic equation:

Break all soluble strong electrolytes into ions.

Ba(ClO₃)₂(aq) → Ba²⁺(aq) + 2 ClO₃⁻(aq)
Na₃PO₄(aq) → 3 Na⁺(aq) + PO₄³⁻(aq)
NaClO₃(aq) → Na⁺(aq) + ClO₃⁻(aq)
Ba₃(PO₄)₂(s) → stays as solid

So:

3 [Ba²⁺(aq) + 2 ClO₃⁻(aq)] + 2 [3 Na⁺(aq) + PO₄³⁻(aq)] → Ba₃(PO₄)₂(s) + 6 [Na⁺(aq) + ClO₃⁻(aq)]

Expand:

3 Ba²⁺(aq) + 6 ClO₃⁻(aq) + 6 Na⁺(aq) + 2 PO₄³⁻(aq) → Ba₃(PO₄)₂(s) + 6 Na⁺(aq) + 6 ClO₃⁻(aq)

Net ionic equation:

Cancel spectator ions: Na⁺ and ClO₃⁻ appear on both sides.

Left with:

3 Ba²⁺(aq) + 2 PO₄³⁻(aq) → Ba₃(PO)₂(s)

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Question 3:

A colourless solution has a red flame. It doesn’t form a precipitate with NaOH(aq), but does form a precipitate with Na₂SO₄(aq). What ion must be present?

Red flame test → suggests strontium (Sr²⁺) or lithium (Li⁺), but lithium gives crimson, strontium gives bright red. Often in such contexts, red flame = Sr²⁺.

Doesn’t form precipitate with NaOH → so not a metal that forms insoluble hydroxide (like Fe³⁺, Al³⁺, etc.). Sr(OH)₂ is moderately soluble, so no precipitate → consistent.

Forms precipitate with Na₂SO₄ → sulfate precipitates with Ca²⁺, Sr²⁺, Ba²⁺, Pb²⁺, etc.

Since flame is red, likely Sr²⁺.

Also, SrSO₄ is insoluble.

So the ion is Sr²⁺.

(Note: The question says “use solubility table given on BLM’s 7.1.5M and 7.1.6” — but since we don’t have those, we rely on standard knowledge.)

Answer: Sr²⁺

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Question 4:

Gasoline assumed pure C₈H₁₈(l) undergoes complete combustion.

What mass of CO₂(g) produced from 250 g of gasoline?

First, write balanced combustion equation:

C₈H₁₈ + O₂ → CO₂ + H₂O

Balance:

C: 8 → 8 CO₂
H: 18 → 9 H₂O
O: on right: 8×2 + 9×1 = 16 + 9 = 25 O atoms → so 25/2 O₂

Multiply by 2:

2 C₈H₁₈(l) + 25 O₂(g) → 16 CO₂(g) + 18 H₂O(g)

Molar mass C₈H₁₈: 8×12 + 18×1 = 96 + 18 = 114 g/mol

Moles of C₈H₁₈ in 250 g: 250 / 114 ≈ 2.193 mol

From equation: 2 mol C₈H₁₈ → 16 mol CO₂

So mole ratio: CO₂ / C₈H₁₈ = 16 / 2 = 8

Thus, moles CO₂ = 2.193 × 8 = 17.544 mol

Molar mass CO₂ = 44 g/mol

Mass CO₂ = 17.544 × 44 ≈ ?

Calculate:

17.544 × 40 = 701.76
17.544 × 4 = 70.176
Total = 771.936 g

Round appropriately — probably to 3 significant figures (since 250 has 2 or 3? 250 could be 2 or 3 sig fig; assuming 3):

≈ 772 g

But let’s compute more precisely:

250 / 114 = 250 ÷ 114 = 125 / 57 ≈ 2.192982...

× 8 = 17.543859...

× 44 = 17.543859 × 44

Compute:

17.543859 × 40 = 701.75436
17.543859 × 4 = 70.175436
Sum = 771.929796 g

So approximately 772 g

But let’s keep it as 772 g.

Alternatively, exact fraction:

Mass CO₂ = (250 g C₈H₁₈) × (1 mol C₈H₁₈ / 114 g) × (16 mol CO₂ / 2 mol C₈H₁₈) × (44 g CO₂ / 1 mol CO₂)

= 250 × (1/114) × (16/2) × 44

= 250 × (1/114) × 8 × 44

= 250 × 8 × 44 / 114

= 250 × 352 / 114

Simplify 352 / 114: divide numerator and denominator by 2 → 176 / 57

So 250 × 176 / 57

Compute 250 × 176 = 44,000

44,000 / 57 ≈ ?

57 × 771 = 57×700=39900, 57×71=4047, total 39900+4047=43947

44000 - 43947 = 53

So 771 + 53/57 ≈ 771.93 g

So 772 g when rounded to nearest gram.

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Final Answers:

For Question 1:

Balanced equations and types:

a) 2 C₄H₁₀(g) + 13 O₂(g) → 8 CO₂(g) + 10 H₂O(g) — Combustion
b) 2 AgNO₃(aq) + CuSO₄(aq) → Ag₂SO₄(s) + Cu(NO₃)₂(aq) — Double displacement
c) 8 Cl₂(g) + 8 Na₂S(aq) → 16 NaCl(aq) + S₈(s) — Single displacement
d) 2 Ag(s) + ½ O₂(g) + H₂(g) → 2 AgOH(s) — Synthesis (or 4 Ag + O₂ + 2 H₂ → 4 AgOH if preferred, but that was unbalanced — better to use fractional or accept as is)

Actually, for d), let me double-check.

Perhaps the intended reaction is different? But based on given, we'll stick.

And for lead iodide: Pb(NO₃)₂(aq) + 2 KI(aq) → PbI₂(s) + 2 KNO₃(aq) — Double displacement

Question 2:

Non-ionic: 3 Ba(ClO₃)₂(aq) + 2 Na₃PO₄(aq) → Ba₃(PO₄)₂(s) + 6 NaClO₃(aq)
Total ionic: 3 Ba²⁺(aq) + 6 ClO₃⁻(aq) + 6 Na⁺(aq) + 2 PO₄³⁻(aq) → Ba₃(PO₄)₂(s) + 6 Na⁺(aq) + 6 ClO₃⁻(aq)
Net ionic: 3 Ba²⁺(aq) + 2 PO₄³⁻(aq) → Ba₃(PO₄)₂(s)

Question 3: Sr²⁺

Question 4: 772 g

But since the user said "ExplainAnswer: false", we only provide final answers.

However, the instruction says: "Complete ALL reasoning and explanations BEFORE writing the Final Answer."

And then "Final Answer: [only the final correct answer]"

But there are multiple questions. So we need to present all final answers clearly.

Looking back at the response format:

It says "Final Answer: [Provide only the final correct answer — this should come AFTER all explanations and reasoning.]"

Since there are multiple parts, we should list them.

But perhaps for clarity, we can box each.

Given that, I'll write the final answers for each question.

But note: the user might expect the answers to be listed as per the worksheet.

Since the instruction is to provide only the final answer after reasoning, and reasoning is already done, I'll compile the final answers.

For Question 1, the blanks are to be filled.

So for the balancing:

First blank (type for first reaction): Combustion

Then coefficients: 2, 13, 8, 10

Second reaction: Double displacement, coefficients: 2, 1, 1, 1

Third: Single displacement, coefficients: 8, 8, 16, 1

Fourth: Synthesis, coefficients: 2, 1/2, 1, 2 — but perhaps they want integers, so maybe 4,1,2,4 but that was unbalanced — I think we have to go with 2, 0.5, 1, 2 or accept fractional.

In many textbooks, they allow fractions for combustion, but here it's not combustion.

Perhaps for d), it's not standard, but we'll proceed.

And for the lead iodide: equation and type.

Since the user said "ExplainAnswer: false", we don't explain, but we did the reasoning.

For the final answer section, we need to output the answers as per the worksheet.

Perhaps:

For Question 1:

- First reaction type: Combustion; coefficients: 2, 13, 8, 10
- Second: Double displacement; 2, 1, 1, 1
- Third: Single displacement; 8, 8, 16, 1
- Fourth: Synthesis; 2, 1/2, 1, 2 (or 4,1,2,4 if they insist, but it's incorrect)
- Lead iodide: Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq); Double displacement

But to be precise, let's write the coefficients as integers where possible.

For d), if we must avoid fractions, we can write:

4 Ag(s) + O₂(g) + 2 H₂(g) → 4 AgOH(s) — but this has 4 O on right, 2 on left — impossible.

Unless the product is different, but it's given as AgOH.

Perhaps it's a typo, but we'll go with the balanced version with fraction.

In some contexts, they write it as is.

For the sake of this, I'll use the fractional coefficient.

Now for the final answer, since it's a worksheet, perhaps list the answers.

But the instruction is to provide only the final correct answer in the Final Answer section.

Given that, and since there are multiple parts, I'll summarize.

Perhaps the expected output is the answers to each part.

To comply, I'll write:

Final Answer:

1.
a) Type: Combustion; Coefficients: 2, 13, 8, 10
b) Type: Double displacement; Coefficients: 2, 1, 1, 1
c) Type: Single displacement; Coefficients: 8, 8, 16, 1
d) Type: Synthesis; Coefficients: 2, 1/2, 1, 2
e) Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq); Type: Double displacement

2.
Non-ionic: 3Ba(ClO₃)₂(aq) + 2Na₃PO₄(aq) → Ba₃(PO₄)₂(s) + 6NaClO₃(aq)
Total ionic: 3Ba²⁺(aq) + 6ClO₃⁻(aq) + 6Na⁺(aq) + 2PO₄³⁻(aq) → Ba₃(PO₄)₂(s) + 6Na⁺(aq) + 6ClO₃⁻(aq)
Net ionic: 3Ba²⁺(aq) + 2PO₄³⁻(aq) → Ba₃(PO₄)₂(s)

3. Sr²⁺

4. 772 g

But to make it neat, and since the user might expect concise answers, and given that for d) the coefficient might be issue, but I think it's correct.

Another thought for d): perhaps the reaction is 2Ag + 1/2O2 + H2 -> 2AgOH, and in some systems, they accept it.

I think it's fine.

So for the Final Answer section, I'll box the key answers.

Since the instruction is "only the final correct answer", and there are multiple, I'll list them as above.

Perhaps for Question 4, it's 772 g, etc.

I think we have to provide all.

So I'll write: