Chemistry worksheet with problems on balancing equations and calculating theoretical and percent yields.

Let's go through each problem step by step and verify the solutions. I'll solve them thoroughly, explain the reasoning, and confirm whether the given answers are correct.

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Problem 1:

Given equation:
$$ \text{K}_2\text{PtCl}_4 + \text{NH}_3 \longrightarrow \text{Pt(NH}_3)_2\text{Cl}_2 + \text{KCl} $$

#### a) Balance the equation

We need to balance atoms on both sides.

- Pt: 1 on each side → balanced.
- Cl: 4 on left (from K₂PtCl₄), 2 in Pt(NH₃)₂Cl₂ and 1 in each KCl → so we need 2 KCl to account for 2 more Cl.
- K: 2 on left → so we need 2 KCl on right → that’s good.
- N: 2 NH₃ needed to make Pt(NH₃)₂Cl₂ → so 2 NH₃ on left.
- H: 6 H from 2 NH₃ → matches with 6 H in Pt(NH₃)₂Cl₂.

So:

$$
\boxed{\text{K}_2\text{PtCl}_4 + 2\text{NH}_3 \longrightarrow \text{Pt(NH}_3)_2\text{Cl}_2 + 2\text{KCl}}
$$

✔ Balanced.

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#### b) Theoretical yield of KCl starting with 34.5 g of NH₃

From the balanced equation:
- 2 mol NH₃ → 2 mol KCl → 1:1 molar ratio between NH₃ and KCl.

Step 1: Moles of NH₃
Molar mass of NH₃ = 14 + 3(1) = 17 g/mol
Moles of NH₃ = 34.5 g / 17 g/mol = 2.029 mol

Since 1 mol NH₃ produces 1 mol KCl, moles of KCl = 2.029 mol

Step 2: Mass of KCl
Molar mass of KCl = 39 + 35.5 = 74.5 g/mol
Mass = 2.029 mol × 74.5 g/mol ≈ 151.2 g

✔ So theoretical yield of KCl is 151 g (rounded). ✔️

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#### c) Percent yield if you isolate 76.4 g of Pt(NH₃)₂Cl₂

We’re told you start with 34.5 g of NH₃ and isolate 76.4 g of Pt(NH₃)₂Cl₂.

We need to find theoretical yield of Pt(NH₃)₂Cl₂ based on NH₃.

From balanced equation:
- 2 mol NH₃ → 1 mol Pt(NH₃)₂Cl₂

Moles of NH₃ = 34.5 / 17 = 2.029 mol
So moles of Pt(NH₃)₂Cl₂ = 2.029 / 2 = 1.0145 mol

Molar mass of Pt(NH₃)₂Cl₂:
- Pt = 195 g/mol
- N₂ = 2×14 = 28
- H₆ = 6×1 = 6
- Cl₂ = 2×35.5 = 71
Total = 195 + 28 + 6 + 71 = 300 g/mol

Theoretical mass = 1.0145 mol × 300 g/mol = 304.35 g

But the answer says 303.9 g, which is very close — likely due to rounding.

Actual yield = 76.4 g

Percent yield = (actual / theoretical) × 100 = (76.4 / 303.9) × 100 ≈ 25.14%

✔ So percent yield = 25.14% ✔️

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Problem 2:

$$
\text{H}_3\text{PO}_4 + 3\text{KOH} \longrightarrow \text{K}_3\text{PO}_4 + 3\text{H}_2\text{O}
$$

Given: 49.0 g H₃PO₄ reacts with excess KOH → isolate 49.0 g K₃PO₄

Find percent yield.

Step 1: Molar masses
- H₃PO₄ = 3(1) + 31 + 4(16) = 3 + 31 + 64 = 98 g/mol
- K₃PO₄ = 3(39) + 31 + 4(16) = 117 + 31 + 64 = 212 g/mol

Step 2: Moles of H₃PO₄
= 49.0 / 98 = 0.5 mol

From equation: 1 mol H₃PO₄ → 1 mol K₃PO₄ → so theoretical moles of K₃PO₄ = 0.5 mol

Theoretical mass = 0.5 × 212 = 106.0 g

Actual yield = 49.0 g

Percent yield = (49.0 / 106.0) × 100 ≈ 46.23%

Answer says 46.17% — slight difference due to rounding?

Wait: let’s check if it's exact.

49.0 / 106.0 = 0.46226... → ×100 = 46.23%

But answer says 46.17% — possibly a typo or different molar mass used?

Let’s double-check molar masses:

- K = 39.1 → 3×39.1 = 117.3
- P = 30.97
- O = 16.00 → 4×16 = 64
- Total K₃PO₄ = 117.3 + 30.97 + 64 = 212.27 g/mol

Then theoretical yield = 0.5 × 212.27 = 106.135 g

Percent yield = 49.0 / 106.135 ≈ 46.17% ✔

So yes, with precise molar masses, it's 46.17% ✔️

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Problem 3:

$$
\text{Al}_2(\text{SO}_4)_3 + 6\text{NaOH} \longrightarrow 3\text{Na}_2\text{SO}_4 + 2\text{Al(OH)}_3
$$

Start with 389.4 g Al₂(SO₄)₃, isolate 212.4 g Na₂SO₄ → find percent yield.

Step 1: Molar masses
- Al₂(SO₄)₃ = 2(27) + 3(32 + 64) = 54 + 3(96) = 54 + 288 = 342 g/mol
- Na₂SO₄ = 2(23) + 32 + 64 = 46 + 32 + 64 = 142 g/mol

Step 2: Moles of Al₂(SO₄)₃
= 389.4 / 342 ≈ 1.1386 mol

From equation: 1 mol Al₂(SO₄)₃ → 3 mol Na₂SO₄
So theoretical moles Na₂SO₄ = 1.1386 × 3 = 3.4158 mol

Theoretical mass = 3.4158 × 142 ≈ 485.5 g

But answer says 500.6 g — this doesn’t match.

Wait — let’s recalculate molar masses carefully.

Al₂(SO₄)₃:
- Al: 2 × 26.98 = 53.96
- S: 3 × 32.07 = 96.21
- O: 12 × 16.00 = 192.00
- Total = 53.96 + 96.21 + 192.00 = 342.17 g/mol

Moles = 389.4 / 342.17 ≈ 1.138 mol

→ 3 × 1.138 = 3.414 mol Na₂SO₄

Na₂SO₄ molar mass:
- Na: 2 × 22.99 = 45.98
- S: 32.07
- O: 64.00 → total = 142.05 g/mol

Theoretical mass = 3.414 × 142.05 ≈ 485.4 g

But the answer says 500.6 g — that’s not matching.

Wait — maybe they used different stoichiometry?

Check: Al₂(SO₄)₃ → 3 Na₂SO₄ → yes.

Wait — perhaps the given answer is wrong?

But let’s check what would give 500.6 g.

Suppose theoretical yield = 500.6 g → moles = 500.6 / 142.05 ≈ 3.524 mol

Then moles of Al₂(SO₄)₃ = 3.524 / 3 = 1.1747 mol

Mass = 1.1747 × 342.17 ≈ 401.6 g — but we have only 389.4 g → too high.

So something is off.

Wait — perhaps I made a mistake in the reaction?

No: Al₂(SO₄)₃ + 6NaOH → 3Na₂SO₄ + 2Al(OH)₃ — correct.

Let me recalculate with approximate values:

Use:
- Al₂(SO₄)₃ = 342 g/mol
- 389.4 / 342 = 1.1386 mol
- Produces 3 × 1.1386 = 3.4158 mol Na₂SO₄
- Na₂SO₄ = 142 g/mol → 3.4158 × 142 = 485.5 g

But the answer says 500.6 g — this is inconsistent.

Wait — could it be a typo? Let's see if Al₂(SO₄)₃ was meant to be something else?

Alternatively, maybe the answer is incorrect.

But let's check the percent yield:

They say actual = 212.4 g, theoretical = 500.6 g → percent = (212.4 / 500.6) × 100 ≈ 42.43%

Which matches their stated percent yield.

So they must have calculated theoretical yield as 500.6 g.

How?

Try:

Suppose molar mass of Al₂(SO₄)₃ = 389.4 / x = moles

Want theoretical Na₂SO₄ = 500.6 g → moles = 500.6 / 142 = 3.525 mol

So moles Al₂(SO₄)₃ = 3.525 / 3 = 1.175 mol

Mass = 1.175 × ? = 389.4 → so molar mass = 389.4 / 1.175 ≈ 331.4 g/mol

But actual is ~342 — so not matching.

Wait — unless the compound is different?

Wait — Al₂(SO₄)₃ is correct.

Wait — maybe they used Al₂(SO₄)₃ as 342, but did:

389.4 g / 342 g/mol = 1.1386 mol

→ 3 × 1.1386 = 3.4158 mol Na₂SO₄

→ 3.4158 × 142 = 485.5 g

But they claim 500.6 g — that’s about 3% higher.

Unless they used different molar mass.

Wait — maybe Na₂SO₄ is taken as 142, but perhaps they used 142.04?

No.

Wait — perhaps there's a mistake in the problem statement?

Alternatively, maybe the reaction is different?

Wait — no, it's standard.

But look: 389.4 g Al₂(SO₄)₃ → how much Na₂SO₄?

Let’s do exact calculation:

- Al₂(SO₄)₃: 2×27 + 3×(32 + 64) = 54 + 3×96 = 54 + 288 = 342 g/mol → correct.
- 389.4 / 342 = 1.1386 mol
- → 3 × 1.1386 = 3.4158 mol Na₂SO₄
- Na₂SO₄: 2×23 + 32 + 64 = 46+32+64=142 g/mol
- 3.4158 × 142 = 485.5 g

But answer says 500.6 g → that’s not possible.

Wait — unless the compound is Al₂(SO₄)₃·18H₂O or hydrated?

But no — not indicated.

Alternatively, maybe it's Al₂(SO₄)₃ but they used wrong molar mass?

Or perhaps the answer is wrong.

But wait — let's check the percent yield:

They say actual = 212.4 g, theoretical = 500.6 g → % = 212.4 / 500.6 × 100 = 42.43%

But if theoretical is 485.5 g → % = 212.4 / 485.5 ≈ 43.77%

Not matching.

So either the theoretical yield is wrong, or the actual is wrong.

But the actual is given as 212.4 g.

Wait — perhaps the balanced equation is wrong?

No — Al₂(SO₄)₃ + 6NaOH → 3Na₂SO₄ + 2Al(OH)₃ — correct.

Wait — unless it's Al₂(SO₄)₃ producing only one Na₂SO₄? No.

Wait — 3 Na₂SO₄ per Al₂(SO₄)₃ — yes.

So unless the mass of Al₂(SO₄)₃ is wrong, or molar mass is wrong.

Wait — let’s try using precise atomic masses:

- Al: 26.98 → 2×26.98 = 53.96
- S: 32.06 → 3×32.06 = 96.18
- O: 16.00 → 12×16.00 = 192.00
- Total Al₂(SO₄)₃ = 53.96 + 96.18 + 192.00 = 342.14 g/mol

389.4 / 342.14 ≈ 1.138 mol

→ 3.414 mol Na₂SO₄

Na₂SO₄:
- Na: 22.99 × 2 = 45.98
- S: 32.06
- O: 64.00 → total = 142.04 g/mol

Theoretical mass = 3.414 × 142.04 ≈ 485.4 g

Still not 500.6.

So answer provided (500.6 g) seems incorrect.

But let's suppose they meant 389.4 g of something else?

Alternatively, maybe the compound is K₂SO₄? No.

Wait — perhaps it's Al₂(SO₄)₃ but they used 342 g/mol, but then:

389.4 / 342 = 1.1386 mol

→ 3 × 1.1386 = 3.4158 mol Na₂SO₄

→ 3.4158 × 142 = 485.5 g

But they say 500.6 g — still off.

Wait — maybe the actual product is not Na₂SO₄, but something else?

No — it says "isolate 212.4 g of Na₂SO₄"

Wait — unless the reaction is different?

Wait — is it possible that the equation is unbalanced?

No — it's balanced.

Wait — maybe it's Al₂(SO₄)₃ + 6NaOH → 3Na₂SO₄ + 2Al(OH)₃ — yes.

I think the given answer is wrong.

But let's move on — perhaps we’ll come back.

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Problem 4:

$$
\text{Al(OH)}_3(s) + 3\text{HCl}(aq) \longrightarrow \text{AlCl}_3(aq) + 3\text{H}_2\text{O}(l)
$$

Start with 50.3 g Al(OH)₃, isolate 39.5 g AlCl₃ → find percent yield.

Step 1: Molar masses
- Al(OH)₃ = 27 + 3(16 + 1) = 27 + 3(17) = 27 + 51 = 78 g/mol
- AlCl₃ = 27 + 3×35.5 = 27 + 106.5 = 133.5 g/mol

Step 2: Moles of Al(OH)₃
= 50.3 / 78 ≈ 0.645 mol

From equation: 1 mol Al(OH)₃ → 1 mol AlCl₃ → so theoretical moles = 0.645 mol

Theoretical mass = 0.645 × 133.5 ≈ 86.0 g

✔ Matches given: 86.0 g

Actual = 39.5 g

Percent yield = (39.5 / 86.0) × 100 ≈ 45.93%

✔ Matches given: 45.93% ✔️

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Problem 5:

$$
\text{K}_2\text{CO}_3 + \text{HCl} \longrightarrow \text{H}_2\text{O} + \text{CO}_2 + \text{KCl}
$$

#### a) Balance the equation

Left: K₂CO₃ + HCl → Right: H₂O + CO₂ + KCl

K: 2 on left → need 2 KCl on right
Cl: 2 on right → need 2 HCl on left
H: 2 HCl → 2 H → need 1 H₂O
O: CO₃ → CO₂ + H₂O → balanced

So:

$$
\text{K}_2\text{CO}_3 + 2\text{HCl} \longrightarrow \text{H}_2\text{O} + \text{CO}_2 + 2\text{KCl}
$$

✔ Balanced.

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#### b) Theoretical yield of KCl from 34.5 g K₂CO₃

Molar mass K₂CO₃ = 2(39) + 12 + 3(16) = 78 + 12 + 48 = 138 g/mol

Moles = 34.5 / 138 = 0.25 mol

From equation: 1 mol K₂CO₃ → 2 mol KCl → so 0.5 mol KCl

Molar mass KCl = 74.5 g/mol → mass = 0.5 × 74.5 = 37.25 g

Round to 37.2 g ✔

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#### c) Starting with 34.5 g K₂CO₃, isolate 34 g H₂O → percent yield?

From earlier: 1 mol K₂CO₃ → 1 mol H₂O

Moles K₂CO₃ = 0.25 mol → theoretical moles H₂O = 0.25 mol

Molar mass H₂O = 18 g/mol → theoretical mass = 0.25 × 18 = 4.5 g

But you isolated 34 g? That can't be — 34 g is way more than 4.5 g.

Wait — this is impossible.

But the answer says 76% — so perhaps 3.4 g?

Ah! Likely typo: 3.4 g of H₂O, not 34 g.

Because:

If actual = 3.4 g

Theoretical = 4.5 g

Percent yield = (3.4 / 4.5) × 100 ≈ 75.56% ≈ 76%

✔ So probably 3.4 g, not 34 g.

Thus, the problem likely has a typo.

So assuming 3.4 g of H₂O isolated → percent yield = 76% ✔️

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Problem 6:

$$
\text{H}_2\text{SO}_4 + \text{Ba(OH)}_2 \longrightarrow \text{BaSO}_4 + \text{H}_2\text{O}
$$

Balance:

H₂SO₄ + Ba(OH)₂ → BaSO₄ + 2H₂O

Yes.

a) 98.0 g H₂SO₄ reacted with excess Ba(OH)₂ → isolate 213.7 g BaSO₄ → find percent yield.

Step 1: Molar masses
- H₂SO₄ = 2(1) + 32 + 64 = 98 g/mol
- BaSO₄ = 137 + 32 + 64 = 233 g/mol

Step 2: Moles of H₂SO₄
= 98.0 / 98 = 1.00 mol

From equation: 1 mol H₂SO₄ → 1 mol BaSO₄ → so theoretical = 1.00 mol

Mass = 1.00 × 233 = 233 g

Actual = 213.7 g

Percent yield = (213.7 / 233) × 100 ≈ 91.72% → rounds to 91.7%

✔ Matches given: 91.7% ✔️

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Summary of Corrections & Verifications:

| Problem | Given Answer | Correct? | Notes |
|--------|--------------|---------|-------|
| 1a | Balanced | ✔ | Correct |
| 1b | 151 g KCl | ✔ | Correct |
| 1c | 25.14% | ✔ | Correct |
| 2 | 46.17% | ✔ | Correct (with precise molar masses) |
| 3 | Theoretical = 500.6 g, 42.43% | ✘ | Incorrect — should be ~485.5 g, ~43.77% |
| 4 | 86.0 g, 45.93% | ✔ | Correct |
| 5a | Balanced | ✔ | Correct |
| 5b | 37.2 g | ✔ | Correct |
| 5c | 76% | ✔ | But actual yield should be 3.4 g, not 34 g |
| 6a | 91.7% | ✔ | Correct |

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Final Answers:

Only Problem 3 has an error.

Corrected Problem 3:

- Molar mass Al₂(SO₄)₃ = 342 g/mol
- Moles = 389.4 / 342 = 1.1386 mol
- → 3 × 1.1386 = 3.4158 mol Na₂SO₄
- Mass = 3.4158 × 142 = 485.5 g theoretical yield
- Actual = 212.4 g
- Percent yield = (212.4 / 485.5) × 100 ≈ 43.77%

But the given answer says 500.6 g and 42.43% — likely a mistake in the problem or answer key.

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✔ All other answers are correct. Only Problem 3 has an error.

Let me know if you want the corrected version written out.