Limiting Reactant-Theoretical Yield and Percent Yield Worksheet ... - Free Printable

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Problem 1:

Reaction:
$$
\text{C}_2\text{H}_4 + 3 \text{O}_2 \rightarrow 2 \text{CO}_2 + 2 \text{H}_2\text{O}
$$

Given:
- 1.0 mol C₂H₄
- 3.0 mol O₂

Question: What is the limiting reactant? How many moles of CO₂ are produced?

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#### Step 1: Determine the stoichiometric ratio

From the balanced equation:
- 1 mol C₂H₄ requires 3 mol O₂

So for 1.0 mol C₂H₄, we need:
$$
1.0 \times 3 = 3.0 \text{ mol O}_2
$$

We have exactly 3.0 mol O₂ → Perfect match!

✔ So neither reactant is in excess — they are in stoichiometric proportions.

Thus, both are completely consumed, so neither is in excess, but both are limiting (in this case, they are equally limiting).

But usually, we say both are limiting or that the reaction goes to completion with no excess.

#### Step 2: Moles of CO₂ produced

From the equation:
- 1 mol C₂H₄ produces 2 mol CO₂
→ So 1.0 mol C₂H₄ produces:
$$
1.0 \times 2 = 2.0 \text{ mol CO}_2
$$

✔ Answer:
- Limiting reactant: Both C₂H₄ and O₂ are limiting (in stoichiometric ratio)
- Moles of CO₂ produced: 2.0 mol

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Problem 2:

Reaction:
$$
\text{SnCl}_2 + 2 \text{AgNO}_3 \rightarrow \text{Sn(NO}_3)_2 + 2 \text{AgCl}
$$

Given:
- 10.0 g SnCl₂
- 10.0 g AgNO₃

Find: Limiting reactant and theoretical yield of AgCl.

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#### Step 1: Molar masses

- SnCl₂: Sn = 118.7, Cl = 35.5 × 2 = 71 → 189.7 g/mol
- AgNO₃: Ag = 107.9, N = 14, O = 16×3 = 48 → 169.9 g/mol
- AgCl: Ag = 107.9, Cl = 35.5 → 143.4 g/mol

#### Step 2: Moles of each reactant

- Moles of SnCl₂ = $ \frac{10.0}{189.7} \approx 0.0527 $ mol
- Moles of AgNO₃ = $ \frac{10.0}{169.9} \approx 0.0589 $ mol

#### Step 3: Use stoichiometry

From the equation:
- 1 mol SnCl₂ reacts with 2 mol AgNO₃

So, 0.0527 mol SnCl₂ would require:
$$
0.0527 \times 2 = 0.1054 \text{ mol AgNO}_3
$$

But we only have 0.0589 mol AgNO₃, which is less than required → AgNO₃ is limiting

Alternatively, check how much SnCl₂ is needed for 0.0589 mol AgNO₃:

$$
\frac{0.0589}{2} = 0.02945 \text{ mol SnCl}_2
$$

We have 0.0527 mol SnCl₂ → more than enough → so AgNO₃ is limiting

✔ Limiting reactant: AgNO₃

#### Step 4: Theoretical yield of AgCl

From the equation:
- 2 mol AgNO₃ → 2 mol AgCl → 1:1 ratio

So 0.0589 mol AgNO₃ → 0.0589 mol AgCl

Mass of AgCl = $ 0.0589 \times 143.4 \approx 8.46 $ g

✔ Theoretical yield of AgCl: 8.46 g

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Problem 3:

Reaction:
$$
3 \text{TiO}_2 + 4 \text{C} + 6 \text{Cl}_2 \rightarrow 3 \text{TiCl}_4 + 2 \text{CO}_2 + 2 \text{CO}
$$

Given:
- 1.0 kg TiO₂ = 1000 g
- 1.0 kg C = 1000 g
- 1.0 kg Cl₂ = 1000 g

Find: Limiting reactant and theoretical yield of TiCl₄

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#### Step 1: Molar masses

- TiO₂: Ti = 47.9, O = 16×2 = 32 → 79.9 g/mol
- C: 12.0 g/mol
- Cl₂: 71.0 g/mol
- TiCl₄: Ti = 47.9, Cl = 35.5×4 = 142 → 189.9 g/mol

#### Step 2: Moles of each

- TiO₂: $ \frac{1000}{79.9} \approx 12.52 $ mol
- C: $ \frac{1000}{12.0} = 83.33 $ mol
- Cl₂: $ \frac{1000}{71.0} \approx 14.08 $ mol

#### Step 3: Stoichiometric ratios

From the equation:
- 3 mol TiO₂ : 4 mol C : 6 mol Cl₂

Let’s find how much of each reactant is needed based on others.

##### Check based on TiO₂:

3 mol TiO₂ → needs 4 mol C and 6 mol Cl₂

So 12.52 mol TiO₂ needs:
- C: $ \frac{4}{3} \times 12.52 = 16.69 $ mol
- Cl₂: $ \frac{6}{3} \times 12.52 = 25.04 $ mol

We have:
- C: 83.33 mol → plenty
- Cl₂: 14.08 mol → not enough → Cl₂ is limiting?

Wait, let's test all.

##### Check based on Cl₂:

6 mol Cl₂ → needs 3 mol TiO₂ and 4 mol C

So 14.08 mol Cl₂ needs:
- TiO₂: $ \frac{3}{6} \times 14.08 = 7.04 $ mol
- C: $ \frac{4}{6} \times 14.08 = 9.39 $ mol

We have:
- TiO₂: 12.52 mol → more than enough
- C: 83.33 mol → more than enough

So Cl₂ is limiting?

Wait — let’s check if any other could be limiting.

##### Check based on C:

4 mol C → needs 3 mol TiO₂ and 6 mol Cl₂

83.33 mol C → needs:
- TiO₂: $ \frac{3}{4} \times 83.33 = 62.5 $ mol → we only have 12.52 mol → not enough TiO₂
- Cl₂: $ \frac{6}{4} \times 83.33 = 125.0 $ mol → we only have 14.08 mol → not enough

So C is not limiting because it would require too much TiO₂ and Cl₂.

Now compare:

- Cl₂ is sufficient for 7.04 mol TiO₂ (we have 12.52 → ok)
- Cl₂ is sufficient for 9.39 mol C (we have 83.33 → ok)

But from Cl₂: 14.08 mol → can produce how much TiCl₄?

From equation:
- 6 mol Cl₂ → 3 mol TiCl₄ → ratio = 0.5 mol TiCl₄ per mol Cl₂

So:
- TiCl₄ = $ 14.08 \times 0.5 = 7.04 $ mol

Now check how much TiCl₄ could be made from TiO₂:
- 3 mol TiO₂ → 3 mol TiCl₄ → 1:1
- 12.52 mol TiO₂ → 12.52 mol TiCl₄ → but limited by Cl₂

And from C:
- 4 mol C → 3 mol TiCl₄ → ratio = 0.75 mol TiCl₄ per mol C
- 83.33 mol C → $ 83.33 \times 0.75 = 62.5 $ mol TiCl₄ → way more than possible

So Cl₂ limits the reaction.

✔ Limiting reactant: Cl₂

#### Step 4: Theoretical yield of TiCl₄

From above:
- 6 mol Cl₂ → 3 mol TiCl₄ → so 1 mol Cl₂ → 0.5 mol TiCl₄
- 14.08 mol Cl₂ → $ 14.08 \times 0.5 = 7.04 $ mol TiCl₄

Mass = $ 7.04 \times 189.9 \approx 1336 $ g ≈ 1.34 kg

✔ Theoretical yield of TiCl₄: 1.34 kg

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Problem 4:

Reaction:
$$
\text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2
$$

Given:
- 100.0 g CaCO₃
- 100.0 g HCl
- Actual yield of CO₂ = 45.0 g
- Find percent yield

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#### Step 1: Molar masses

- CaCO₃: Ca = 40, C = 12, O = 48 → 100.0 g/mol
- HCl: 1 + 35.5 = 36.5 g/mol
- CO₂: 12 + 32 = 44.0 g/mol

#### Step 2: Moles

- CaCO₃: $ \frac{100.0}{100.0} = 1.00 $ mol
- HCl: $ \frac{100.0}{36.5} \approx 2.74 $ mol

#### Step 3: Stoichiometry

From equation:
- 1 mol CaCO₃ requires 2 mol HCl

So 1.00 mol CaCO₃ needs 2.00 mol HCl

We have 2.74 mol HCl → more than enough → so CaCO₃ is limiting

HCl is in excess.

#### Step 4: Theoretical yield of CO₂

From equation:
- 1 mol CaCO₃ → 1 mol CO₂
- So 1.00 mol CaCO₃ → 1.00 mol CO₂

Mass = $ 1.00 \times 44.0 = 44.0 $ g

#### Step 5: Percent yield

$$
\text{Percent yield} = \frac{\text{actual}}{\text{theoretical}} \times 100 = \frac{45.0}{44.0} \times 100 \approx 102.3\%
$$

Wait — over 100%? That’s impossible unless there’s error.

But 45.0 / 44.0 = 1.0227 → 102.3% → suggests error in measurement or impurity, but theoretically, it's not possible.

But maybe typo? Let’s double-check.

Wait — actual yield = 45.0 g, theoretical = 44.0 g → yes, over 100%.

But perhaps calculation error?

No — 100.0 g CaCO₃ → 1.00 mol → 1.00 mol CO₂ → 44.0 g → correct.

HCl: 100.0 g / 36.5 = 2.74 mol → 2.74 / 2 = 1.37 mol CaCO₃ needed → but we have only 1.00 mol → so yes, CaCO₃ is limiting.

So theoretical yield is 44.0 g

But actual is 45.0 g → 102.3% yield → impossible unless impurities or mistake.

But perhaps the problem expects us to report it anyway.

✔ Percent yield = 102.3% (but this indicates an experimental error)

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✔ Final Answers:

#### Problem 1:
- Limiting reactant: Both C₂H₄ and O₂ (stoichiometric)
- Moles of CO₂ produced: 2.0 mol

#### Problem 2:
- Limiting reactant: AgNO₃
- Theoretical yield of AgCl: 8.46 g

#### Problem 3:
- Limiting reactant: Cl₂
- Theoretical yield of TiCl₄: 1.34 kg

#### Problem 4:
- Percent yield of CO₂: 102.3% (suggests experimental error)

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