SOLUTION: Chemistry stoichiometry practice problems with answers ... - Free Printable

You’ve uploaded a stoichiometry worksheet with 6 problems, and you’re asking to solve the problem and explain the solution. Since you didn’t specify which problem, I’ll go through each one, verify the given answers, and provide a clear, step-by-step explanation for each — including balancing equations where needed.

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## ✔ Problem 1:
> Calculate the number of moles of oxygen required to react completely with 14.8 mol Al in the following equation:
> 4Al(s) + 3O₂(g) → 2Al₂O₃(s)

✔️ Given Answer: 11.1 mol O₂

📝 Explanation:

This is a simple mole-to-mole conversion using the balanced equation.

From the equation:
4 mol Al : 3 mol O₂

So, set up a proportion:

\[
\text{mol O}_2 = 14.8 \, \text{mol Al} \times \frac{3 \, \text{mol O}_2}{4 \, \text{mol Al}} = \frac{44.4}{4} = 11.1 \, \text{mol O}_2
\]

✔ Correct!

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## ✔ Problem 2:
> Calculate the moles of oxygen produced from the decomposition of 0.61 moles of potassium chlorate (KClO₃).
> 2KClO₃ → 2KCl + 3O₂

✔️ Given Answer: 0.915 mol O₂

📝 Explanation:

From the balanced equation:
2 mol KClO₃ → 3 mol O₂

So,

\[
\text{mol O}_2 = 0.61 \, \text{mol KClO}_3 \times \frac{3 \, \text{mol O}_2}{2 \, \text{mol KClO}_3} = \frac{1.83}{2} = 0.915 \, \text{mol O}_2
\]

✔ Correct!

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## ✔ Problem 3:
> If 20.0 g of zinc react with excess hydrochloric acid, how many grams of zinc chloride (ZnCl₂) are produced?
> Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

*(Note: Equation was not fully balanced in image — but it is already balanced as written.)*

✔️ Given Answer: 41.8 g ZnCl₂

📝 Step-by-Step Solution:

1. Molar Masses:
- Zn: 65.38 g/mol ≈ 65 g/mol (used in worksheet)
- ZnCl₂: 65 + 2×35.45 = 65 + 70.9 = 135.9 g/mol ≈ 136 g/mol

2. Convert mass Zn → moles Zn:
\[
\frac{20.0 \, \text{g Zn}}{65 \, \text{g/mol}} = 0.3077 \, \text{mol Zn}
\]

3. Use mole ratio (1:1 for Zn : ZnCl₂):
→ 0.3077 mol ZnCl₂ produced

4. Convert moles ZnCl₂ → grams:
\[
0.3077 \, \text{mol} \times 136 \, \text{g/mol} = 41.85 \, \text{g} ≈ \boxed{41.8 \, \text{g}}
\]

✔ Correct!

*(Note: Worksheet used approximate molar masses — 65 for Zn, 136 for ZnCl₂ — which is acceptable for high school level.)*

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## ✔ Problem 4:
> How many grams of chlorine gas must be reacted with excess sodium iodide (NaI) if 10.0 grams of sodium chloride (NaCl) are needed?
> 2NaI(aq) + Cl₂(g) → 2NaCl(aq) + I₂(g)

*(Equation was incomplete — needs balancing. It’s already balanced as written.)*

✔️ Given Answer: 6.07 g Cl₂

📝 Step-by-Step Solution:

1. Molar Masses:
- NaCl: 23 + 35.45 = 58.45 g/mol ≈ 58.5 g/mol (used in worksheet)
- Cl₂: 2 × 35.45 = 70.9 g/mol ≈ 71 g/mol

2. Convert mass NaCl → moles NaCl:
\[
\frac{10.0 \, \text{g NaCl}}{58.5 \, \text{g/mol}} = 0.1709 \, \text{mol NaCl}
\]

3. Use mole ratio (2 mol NaCl : 1 mol Cl₂):
\[
\text{mol Cl}_2 = 0.1709 \, \text{mol NaCl} \times \frac{1 \, \text{mol Cl}_2}{2 \, \text{mol NaCl}} = 0.08545 \, \text{mol Cl}_2
\]

4. Convert moles Cl₂ → grams:
\[
0.08545 \, \text{mol} \times 71 \, \text{g/mol} = 6.067 \, \text{g} ≈ \boxed{6.07 \, \text{g}}
\]

✔ Correct!

---

## ✔ Problem 5:
> How many grams of copper are required to replace 4.00 grams of silver nitrate which are dissolved in water?
> Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)

*(Equation was partially written — we balance it: 1 Cu, 2 Ag, 2 N, 6 O on both sides — already balanced.)*

✔️ Given Answer: *(Worksheet shows calculation leading to ~1.5 g Cu — let’s check)*

Wait — the worksheet shows:

> 4.0 g AgNO₃ | 1 mol AgNO₃ | 1 mol Cu | 63.5 g Cu
> | 170 g AgNO₃ | 2 mol AgNO₃ | 1 mol Cu
> → Result: = ? (worksheet says “= 1.5 g Cu” — but let’s calculate)

📝 Step-by-Step Solution:

1. Molar Masses:
- AgNO₃: 107.87 (Ag) + 14.01 (N) + 48 (O×3) = 169.88 g/mol ≈ 170 g/mol
- Cu: 63.5 g/mol

2. Moles of AgNO₃:
\[
\frac{4.00 \, \text{g}}{170 \, \text{g/mol}} = 0.02353 \, \text{mol AgNO}_3
\]

3. Mole Ratio (2 mol AgNO₃ : 1 mol Cu):
\[
\text{mol Cu} = 0.02353 \times \frac{1}{2} = 0.01176 \, \text{mol Cu}
\]

4. Mass of Cu:
\[
0.01176 \, \text{mol} \times 63.5 \, \text{g/mol} = \boxed{0.747 \, \text{g Cu}}
\]

⚠️ Worksheet answer is WRONG! It says 1.5 g, but correct answer is ~0.75 g Cu.

The worksheet likely made an error by using 1:1 ratio instead of 2:1.

✔ Correction: Answer should be 0.75 g Cu

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## ✔ Problem 6:
> How many moles of hydrogen gas are needed to react with 15.1 g chlorine gas to produce hydrogen chloride gas?
> H₂(g) + Cl₂(g) → 2HCl(g)

*(Equation was incomplete — needs balancing. Balanced as written.)*

✔️ Given Answer: 0.21 moles H₂

📝 Step-by-Step Solution:

1. Molar Mass Cl₂: 70.9 g/mol ≈ 71 g/mol (used in worksheet)

2. Moles of Cl₂:
\[
\frac{15.1 \, \text{g}}{71 \, \text{g/mol}} = 0.2127 \, \text{mol Cl}_2
\]

3. Mole Ratio (1:1 for H₂ : Cl₂):
→ 0.2127 mol H₂ needed

4. Rounded to 2 sig figs:
\[
\boxed{0.21 \, \text{mol H}_2}
\]

✔ Correct!

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## 🎯 Final Summary:

| Problem | Correct? | Notes |
|---------|----------|-------|
| 1 | ✔ Yes | Simple mole ratio |
| 2 | ✔ Yes | Mole ratio from decomposition |
| 3 | ✔ Yes | Mass → moles → mass using molar masses |
| 4 | ✔ Yes | Mass → moles → mass using 2:1 ratio |
| 5 | ✘ No | Worksheet error — correct answer is 0.75 g Cu, not 1.5 g |
| 6 | ✔ Yes | Mass → moles → moles using 1:1 ratio |

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## 💡 Tip for Students:
Always double-check:
- Is the chemical equation balanced?
- Are you using the correct mole ratio?
- Are your molar masses accurate?
- Did you convert mass ↔ moles correctly?

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✔ You now have fully explained, verified solutions for all 6 problems — with correction for Problem 5.

Let me know if you’d like this formatted as a printable answer key or want help with more worksheets!