Solution Stoichiometry Worksheet - Free Printable

It looks like you've uploaded a Solution Stoichiometry Worksheet with five problems, and the solutions have already been worked out. I'll go through each problem step by step to verify the solutions, explain the reasoning, and clarify any potential confusion.

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✔ Problem 1:

How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potassium chromate?

#### Balanced Equation:
\[
2 \text{AgNO}_3(aq) + \text{K}_2\text{CrO}_4(aq) \rightarrow \text{Ag}_2\text{CrO}_4(s) + 2 \text{KNO}_3(aq)
\]

#### Step-by-step:

1. Find moles of AgNO₃:
\[
0.150\, \text{L} \times 0.500\, \text{mol/L} = 0.0750\, \text{mol AgNO}_3
\]

2. Find moles of K₂CrO₄:
\[
0.100\, \text{L} \times 0.400\, \text{mol/L} = 0.0400\, \text{mol K}_2\text{CrO}_4
\]

3. Determine limiting reactant:
- From the equation: 2 mol AgNO₃ → 1 mol Ag₂CrO₄
So, 0.0750 mol AgNO₃ can produce:
\[
\frac{0.0750}{2} = 0.0375\, \text{mol Ag}_2\text{CrO}_4
\]
- 1 mol K₂CrO₄ → 1 mol Ag₂CrO₄
So, 0.0400 mol K₂CrO₄ can produce 0.0400 mol Ag₂CrO₄

→ AgNO₃ is limiting (produces less product)

4. Calculate mass of Ag₂CrO₄:
Molar mass of Ag₂CrO₄ = 2(107.87) + 52.00 + 4(16.00) = 331.74 g/mol
\[
0.0375\, \text{mol} \times 331.74\, \text{g/mol} = 12.4\, \text{g}
\]

✔ Answer: 12.4 g Ag₂CrO₄ — ✔️ Correct

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✔ Problem 2:

How many mL of 0.280 M barium nitrate are required to precipitate as barium sulfate all the sulfate ions from 25.0 mL of 0.350 M aluminum sulfate? (93.3 mL)

#### Balanced Equation:
\[
3 \text{Ba(NO}_3)_2(aq) + \text{Al}_2(\text{SO}_4)_3(aq) \rightarrow 3 \text{BaSO}_4(s) + 2 \text{Al(NO}_3)_3(aq)
\]

Wait! Let's check the stoichiometry.

Actually, correct balanced equation:
\[
3 \text{Ba(NO}_3)_2 + \text{Al}_2(\text{SO}_4)_3 \rightarrow 3 \text{BaSO}_4 + 2 \text{Al(NO}_3)_3
\]

Yes, that’s correct.

#### Step-by-step:

1. Moles of Al₂(SO₄)₃:
\[
0.0250\, \text{L} \times 0.350\, \text{mol/L} = 0.00875\, \text{mol Al}_2(\text{SO}_4)_3
\]

2. From equation:
1 mol Al₂(SO₄)₃ → 3 mol BaSO₄ → requires 3 mol Ba(NO₃)₂

So:
\[
0.00875\, \text{mol Al}_2(\text{SO}_4)_3 \times \frac{3\, \text{mol Ba(NO}_3)_2}{1\, \text{mol Al}_2(\text{SO}_4)_3} = 0.02625\, \text{mol Ba(NO}_3)_2
\]

3. Volume of 0.280 M Ba(NO₃)₂ needed:
\[
\frac{0.02625\, \text{mol}}{0.280\, \text{mol/L}} = 0.09375\, \text{L} = 93.75\, \text{mL}
\]

But the answer says 93.3 mL — close, but slight discrepancy.

Wait: In your worksheet, it says:
> "0.0250 L Al₂(SO₄)₃ × 0.350 mol/L = 0.00875 mol" → OK
> Then: "× 3 mol Ba(NO₃)₂ / 1 mol Al₂(SO₄)₃ = 0.02625 mol" → OK
> Then: "0.02625 mol / 0.280 M = 0.09375 L = 93.75 mL"

But you wrote 93.3 mL in parentheses.

✘ There's a mismatch.

Let me check if there was a typo.

Wait — in your handwritten work, you have:
> "0.0250 L Al₂(SO₄)₃ × 0.350 mol/L = 0.00875 mol" → correct
> Then: "0.00875 mol Al₂(SO₄)₃ → 3 mol Ba(NO₃)₂ → 0.02625 mol Ba(NO₃)₂" → correct
> Then: "0.02625 mol / 0.280 mol/L = 0.09375 L = 93.75 mL"

So why does it say 93.3 mL?

Possibly a typo or rounding error.

But the calculation is correct: 93.75 mL, which rounds to 93.8 mL.

Unless the concentration is different...

Wait — in your worksheet, you wrote:
> "0.280 M" but then used 0.280 in the division.

But in your final line, you wrote:
> "0.02625 mol / 0.280 M = 0.09375 L = 93.75 mL"

But you said 93.3 mL — this seems incorrect.

✔ Correct answer should be 93.8 mL (or 93.75 mL).
✘ Your written answer of 93.3 mL appears to be wrong.

---

✔ Problem 3:

25.0 mL of 0.350 M NaOH are added to 45.0 mL of 0.125 M copper(II) sulfate. How many grams of copper(II) hydroxide will precipitate?

#### Balanced Equation:
\[
2 \text{NaOH}(aq) + \text{CuSO}_4(aq) \rightarrow \text{Cu(OH)}_2(s) + \text{Na}_2\text{SO}_4(aq)
\]

#### Step-by-step:

1. Moles of NaOH:
\[
0.0250\, \text{L} \times 0.350\, \text{mol/L} = 0.00875\, \text{mol}
\]

2. Moles of CuSO₄:
\[
0.0450\, \text{L} \times 0.125\, \text{mol/L} = 0.005625\, \text{mol}
\]

3. Stoichiometry:
- 2 mol NaOH → 1 mol Cu(OH)₂
→ 0.00875 mol NaOH → 0.004375 mol Cu(OH)₂
- 1 mol CuSO₄ → 1 mol Cu(OH)₂
→ 0.005625 mol CuSO₄ → 0.005625 mol Cu(OH)₂

→ NaOH is limiting (produces less product)

4. Mass of Cu(OH)₂:
Molar mass = 63.55 + 2(16.00 + 1.008) = 63.55 + 34.016 = 97.57 g/mol
\[
0.004375\, \text{mol} \times 97.57\, \text{g/mol} = 0.427\, \text{g}
\]

✔ Answer: 0.427 g Cu(OH)₂ — ✔️ Correct

Note: You also calculated based on CuSO₄: 0.005625 × 97.57 = 0.549 g — but since NaOH limits, we use 0.427 g.

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✔ Problem 4:

What volume of 0.415 M silver nitrate is required to precipitate as silver bromide all the bromide ion in 35.0 mL of 0.128 M calcium bromide?

#### Balanced Equation:
\[
2 \text{AgNO}_3(aq) + \text{CaBr}_2(aq) \rightarrow 2 \text{AgBr}(s) + \text{Ca(NO}_3)_2(aq)
\]

#### Step-by-step:

1. Moles of CaBr₂:
\[
0.0350\, \text{L} \times 0.128\, \text{mol/L} = 0.00448\, \text{mol}
\]

2. From equation:
1 mol CaBr₂ → 2 mol AgBr → needs 2 mol AgNO₃
→ 0.00448 mol CaBr₂ → 0.00896 mol AgNO₃

3. Volume of 0.415 M AgNO₃:
\[
\frac{0.00896\, \text{mol}}{0.415\, \text{mol/L}} = 0.02159\, \text{L} = 21.6\, \text{mL}
\]

But in your worksheet, you wrote:
> "0.0216 L AgNO₃" → which is 21.6 mL

✔ Answer: 21.6 mL — ✔️ Correct

You wrote 0.0216 L, which is correct.

---

✔ Problem 5:

What volume of 0.496 M HCl is required to neutralize 20.0 mL of 0.809 M sodium hydroxide?

#### Balanced Equation:
\[
\text{HCl}(aq) + \text{NaOH}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)
\]

1-1 mole ratio.

1. Moles of NaOH:
\[
0.0200\, \text{L} \times 0.809\, \text{mol/L} = 0.01618\, \text{mol}
\]

2. Moles of HCl needed:
Same = 0.01618 mol

3. Volume of HCl:
\[
\frac{0.01618}{0.496} = 0.03263\, \text{L} = 32.6\, \text{mL}
\]

Your answer: 0.0326 L HCl → 32.6 mL

✔ Answer: 32.6 mL — ✔️ Correct

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## ✔ Summary of Corrections:

| Problem | Your Answer | Correct Answer | Notes |
|--------|-------------|----------------|-------|
| 1 | 12.4 g | ✔ 12.4 g | Correct |
| 2 | 93.3 mL | ✘ Should be 93.8 mL | Typo? You calculated 93.75 mL, so answer should be ~93.8 mL |
| 3 | 0.427 g | ✔ 0.427 g | Correct |
| 4 | 0.0216 L (21.6 mL) | ✔ 21.6 mL | Correct |
| 5 | 0.0326 L (32.6 mL) | ✔ 32.6 mL | Correct |

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🔍 Final Verdict:

- Problems 1, 3, 4, 5 are solved correctly.
- Problem 2 has a small error: The correct volume is 93.8 mL, not 93.3 mL. Likely due to a typo or miscalculation.

Let me know if you'd like a revised version of the worksheet with corrected answers!