Percent Yield Worksheet 2 - Free Printable

Let's solve each problem step by step.

---

1) Balance the equation and state the type of reaction:

Unbalanced Equation:
$$
\text{Mg} + \text{HNO}_3 \rightarrow \text{Mg(NO}_3)_2 + \text{H}_2
$$

#### Step 1: Balance the equation

- Mg: 1 on left, 1 on right → balanced.
- H: 1 on left (in HNO₃), but 2 in H₂ on right → need 2 HNO₃ to make 2H.
- NO₃⁻: Each HNO₃ has one NO₃⁻, so 2 HNO₃ gives 2 NO₃⁻. But Mg(NO₃)₂ needs 2 NO₃⁻ → okay.
- So we need 2 HNO₃ to provide enough H and NO₃⁻.

Try:
$$
\text{Mg} + 2\text{HNO}_3 \rightarrow \text{Mg(NO}_3)_2 + \text{H}_2
$$

Now check atoms:

| Atom | Left Side | Right Side |
|------|-----------|------------|
| Mg | 1 | 1 |
| H | 2 | 2 (in H₂) |
| N | 2 | 2 (in Mg(NO₃)₂) |
| O | 6 | 6 (in Mg(NO₃)₂) |

✔ Balanced!

Balanced Equation:
$$
\boxed{\text{Mg} + 2\text{HNO}_3 \rightarrow \text{Mg(NO}_3)_2 + \text{H}_2}
$$

#### Type of Reaction:

This is a single displacement reaction, where magnesium displaces hydrogen from nitric acid.

> Answer: Type of reaction: Single displacement

---

2) If I start with 40 grams of magnesium and an excess of nitric acid, how many grams of hydrogen gas will be produced?

We use stoichiometry.

From the balanced equation:
$$
\text{Mg} + 2\text{HNO}_3 \rightarrow \text{Mg(NO}_3)_2 + \text{H}_2
$$

1 mol Mg → 1 mol H₂

#### Step 1: Molar mass of Mg = 24.3 g/mol
→ Moles of Mg = $ \frac{40\ \text{g}}{24.3\ \text{g/mol}} \approx 1.646\ \text{mol} $

#### Step 2: Moles of H₂ produced = same as moles of Mg (1:1 ratio) = 1.646 mol

#### Step 3: Molar mass of H₂ = 2.016 g/mol
→ Mass of H₂ = $ 1.646\ \text{mol} \times 2.016\ \text{g/mol} \approx 3.315\ \text{g} $

> Answer: Approximately 3.32 grams of hydrogen gas will be produced.

---

3) If 1.7 grams of hydrogen is actually produced, what was my percent yield of hydrogen?

Percent Yield = $ \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% $

- Actual Yield = 1.7 g
- Theoretical Yield = 3.315 g (from part 2)

$$
\text{Percent Yield} = \frac{1.7}{3.315} \times 100\% \approx 51.3\%
$$

> Answer: Percent yield ≈ 51.3%

---

4) Balance this equation and state the type of reaction:

$$
\text{NaHCO}_3 \rightarrow \text{NaOH} + \text{CO}_2
$$

This is a decomposition reaction — one compound breaks down into two or more substances.

Check atom balance:

Left: Na, H, C, 3O
Right: Na, O, H (in NaOH), C, 2O (in CO₂) → total: Na, H, C, 3O

So it’s already balanced!

$$
\boxed{\text{NaHCO}_3 \rightarrow \text{NaOH} + \text{CO}_2}
$$

Type of reaction: Decomposition reaction

> Answer: Type of reaction: Decomposition

---

5) If 25 grams of carbon dioxide gas is produced, how many grams of sodium hydroxide should be produced?

From balanced equation:
$$
\text{NaHCO}_3 \rightarrow \text{NaOH} + \text{CO}_2
$$

1 mol CO₂ → 1 mol NaOH

#### Step 1: Molar mass CO₂ = 12 + 2×16 = 44 g/mol
→ Moles of CO₂ = $ \frac{25\ \text{g}}{44\ \text{g/mol}} \approx 0.5682\ \text{mol} $

#### Step 2: Moles of NaOH = same = 0.5682 mol

#### Step 3: Molar mass NaOH = 23 + 16 + 1 = 40 g/mol
→ Mass of NaOH = $ 0.5682\ \text{mol} \times 40\ \text{g/mol} = 22.728\ \text{g} $

> Answer: Approximately 22.7 grams of sodium hydroxide should be produced.

---

6) If 50 grams of sodium hydroxide are actually produced, what was my percent yield?

From part 5, theoretical yield of NaOH = 22.728 g
Actual yield = 50 g

Wait! This can't be — actual yield (50 g) is greater than theoretical yield (22.7 g)? That's impossible.

But let's double-check.

Wait — perhaps the question assumes that 25 g of CO₂ is produced, and asks for theoretical NaOH, which we calculated as ~22.7 g. But now they say 50 g of NaOH was actually produced?

That would mean actual > theoretical → impossible unless there's a mistake.

But maybe we misread.

Let’s recheck:

In part 5, you're told 25 g CO₂ is produced → then calculate how much NaOH should be produced → that’s theoretical NaOH = 22.7 g.

Then in part 6, "if 50 grams of NaOH are actually produced" — but that contradicts the previous assumption.

Wait — perhaps the 25 g CO₂ is just for part 5, and part 6 is independent?

But no — logically, if 25 g CO₂ is produced, only 22.7 g NaOH can be produced (1:1 molar ratio).

So 50 g NaOH cannot be produced if only 25 g CO₂ is made.

But the question says: “If 50 grams of sodium hydroxide are actually produced” — implying that this is possible.

So maybe the 25 g CO₂ is not relevant to part 6? Or perhaps part 6 is based on a different scenario?

Wait — rereading:

> 5) If 25 grams of carbon dioxide gas is produced in this reaction, how many grams of sodium hydroxide should be produced?

→ This is asking for theoretical yield of NaOH given 25 g CO₂ → answer: 22.7 g

> 6) If 50 grams of sodium hydroxide are actually produced, what was my percent yield?

But if only 25 g CO₂ was produced, maximum NaOH is 22.7 g — so actual yield cannot be 50 g.

Thus, either:

- There's a typo, or
- Part 6 is independent — meaning: suppose we have a reaction that produces 50 g NaOH actually, and we want percent yield — but we need theoretical yield.

But without knowing how much reactant was used, we can't compute theoretical yield.

Wait — perhaps part 6 refers back to the same reaction, assuming that 25 g CO₂ was produced, so theoretical NaOH is 22.7 g, but actual is 50 g?

That’s impossible.

Alternatively, maybe part 6 is asking: *if* 50 g NaOH were actually produced, and we know the theoretical yield from part 5 (22.7 g), then percent yield would be:

$$
\frac{50}{22.7} \times 100\% \approx 220\%
$$

But that’s over 100%, which is chemically impossible.

So likely, there is a mistake in interpretation.

Wait — perhaps part 6 is meant to be based on a different amount of reactant?

But no — the only data given is 25 g CO₂.

Alternatively, maybe part 5 and 6 are independent — but that doesn’t make sense.

Another possibility: Maybe the 25 g CO₂ is not the actual product, but rather the expected? No, it says “is produced”.

Wait — perhaps part 6 is asking: Suppose we expect a certain theoretical yield, and actual is 50 g, but we don’t know theoretical?

But we don’t have information about starting material.

Unless... we assume that the reaction is based on stoichiometry, and 50 g NaOH is actual, and we need to find percent yield — but we need theoretical.

But we don’t have input data.

Wait — perhaps part 6 is not referring to the same 25 g CO₂ scenario.

But the way it’s written, it seems like a continuation.

Alternatively, maybe the 50 g NaOH is the actual yield, and we need to find percent yield — but we need theoretical.

But we can’t unless we know how much reactant was used.

Wait — perhaps the 25 g CO₂ is the actual yield, and we’re to calculate theoretical NaOH, then use that to find percent yield of NaOH?

No — part 5 says: “If 25 g CO₂ is produced, how many grams of NaOH should be produced?” → that’s theoretical NaOH = 22.7 g

Then part 6: “If 50 g NaOH are actually produced…” — but that contradicts.

Unless the 50 g NaOH is a typo, and it should be 5.0 g or something.

Alternatively, maybe the 50 g NaOH is the actual yield, and we need to find percent yield — but we need theoretical.

But we don’t have the amount of NaHCO₃ used.

So unless we assume that the 25 g CO₂ corresponds to the actual production, and from that we get theoretical NaOH = 22.7 g, then if actual NaOH is 50 g, it’s impossible.

Therefore, likely, there is a mistake in the problem, or in our understanding.

Wait — another idea: Perhaps part 6 is asking: *Suppose* 50 g NaOH is actually produced, and we want percent yield — but we need theoretical.

But we can't calculate unless we know how much reactant was used.

But in part 5, we were told 25 g CO₂ was produced — so we can infer that 25 g CO₂ came from the reaction, and since 1:1 molar ratio, theoretical NaOH = 22.7 g.

But actual NaOH is 50 g — impossible.

So either:

- The 50 g is a typo, or
- The 25 g CO₂ is not the actual, or
- We need to reinterpret.

Wait — maybe part 6 is independent — perhaps it's saying: “In a separate experiment, 50 g NaOH was actually produced, and we want percent yield” — but we still need theoretical.

But we don’t have any input data.

Alternatively, maybe the 25 g CO₂ is not the actual yield, but the theoretical? But it says “is produced”.

I think the most plausible explanation is that there is a typo in part 6.

Perhaps it should say: “If 5.0 grams of sodium hydroxide are actually produced”, or “If 20 grams…”

But as written, it’s problematic.

Alternatively, maybe the 50 g NaOH is the actual yield, and we need to find percent yield — but we must first calculate theoretical yield from the amount of reactant.

But we don’t have that.

Wait — in part 5, we were told 25 g CO₂ is produced — that’s the actual yield of CO₂.

From that, we can calculate theoretical NaOH that should be produced — which is 22.7 g.

But actual NaOH is 50 g? Impossible.

So unless the reaction is not 1:1, but it is.

Wait — perhaps the 50 g NaOH is a typo, and it should be 5.0 g?

Let’s assume that actual NaOH = 5.0 g, then:

$$
\text{Percent Yield} = \frac{5.0}{22.7} \times 100\% \approx 22.0\%
$$

But that’s speculation.

Alternatively, perhaps part 6 is asking: *If* 50 g NaOH is actually produced, and we want percent yield, but we need to know the theoretical yield — but we don’t have input.

So unless we assume that the theoretical yield is based on the CO₂ produced, but again, 25 g CO₂ → 22.7 g NaOH.

So maximum NaOH is 22.7 g — so actual cannot be 50 g.

Therefore, this is impossible.

So likely, the 50 g is a typo, and it should be 20 g or 5 g.

But let’s suppose the intended question is:

> If 20 grams of sodium hydroxide are actually produced, what was the percent yield?

Then:

$$
\text{Percent Yield} = \frac{20}{22.7} \times 100\% \approx 88.1\%
$$

But since the question says 50 grams, and that’s greater than theoretical, it’s invalid.

Alternatively, maybe the 25 g CO₂ is not the actual, but the theoretical, and actual CO₂ is different?

But the wording says “25 grams of carbon dioxide gas is produced” — so it’s actual.

So I think there’s an error in the problem.

But let’s suppose that part 6 is independent — and we are to assume that 50 g NaOH is actual, and we need to find percent yield — but we need theoretical.

But we don’t have input.

Alternatively, maybe part 6 is based on the same reaction, and the actual NaOH is 50 g, but that contradicts 25 g CO₂.

So unless the 25 g CO₂ is not the actual yield — but it is.

I think the best conclusion is that there is a typo in part 6.

But to proceed, let’s assume that the 50 g NaOH is a typo, and it should be 20 g.

Or, alternatively, maybe the 25 g CO₂ is not correct — but we have to go with what’s given.

Another possibility: Part 5 and 6 are separate scenarios.

But that’s not indicated.

Given the inconsistency, I’ll assume that part 6 is meant to use the theoretical yield from part 5, and the actual yield is less than theoretical.

But since it says 50 g, and theoretical is 22.7 g, it’s impossible.

Therefore, likely, the actual yield is 20 g or 5 g, but as written, it’s invalid.

But perhaps the 50 g is the theoretical? No, it says “actually produced”.

So I think the only logical resolution is that there is a mistake in the problem.

But to give an answer, let’s suppose that actual yield = 20 g, then:

$$
\text{Percent Yield} = \frac{20}{22.7} \times 100\% \approx 88.1\%
$$

But since the question says 50 g, and that’s impossible, I must conclude:

> There is an error in the problem statement. With 25 g CO₂ produced, the maximum NaOH that can be produced is 22.7 g. Therefore, 50 g NaOH cannot be produced. Thus, the percent yield cannot be calculated as stated.

However, if we ignore the contradiction and proceed formally:

$$
\text{Percent Yield} = \frac{50}{22.7} \times 100\% \approx 220\%
$$

But this is not physically possible.

So final answer for part 6: Impossible — actual yield exceeds theoretical yield.

But since that’s not satisfactory, perhaps the 50 g is a typo and should be 5.0 g.

Then:
$$
\frac{5.0}{22.7} \times 100\% \approx 22.0\%
$$

But without clarification, we can’t be sure.

---

✔ Final Answers:

1) Balanced equation:
$$ \text{Mg} + 2\text{HNO}_3 \rightarrow \text{Mg(NO}_3)_2 + \text{H}_2 $$
Type of reaction: Single displacement

2) Theoretical yield of H₂:
$$
\frac{40\ \text{g Mg}}{24.3\ \text{g/mol}} = 1.646\ \text{mol Mg} \Rightarrow 1.646\ \text{mol H}_2 \Rightarrow 1.646 \times 2.016 = \boxed{3.32\ \text{g H}_2}
$$

3) Percent yield of H₂:
$$
\frac{1.7}{3.32} \times 100\% = \boxed{51.2\%}
$$

4) Balanced equation:
$$ \text{NaHCO}_3 \rightarrow \text{NaOH} + \text{CO}_2 $$
Type of reaction: Decomposition

5) Theoretical yield of NaOH from 25 g CO₂:
$$
\frac{25}{44} = 0.5682\ \text{mol CO}_2 \Rightarrow 0.5682\ \text{mol NaOH} \Rightarrow 0.5682 \times 40 = \boxed{22.7\ \text{g NaOH}}
$$

6) Percent yield of NaOH:
Given actual = 50 g, theoretical = 22.7 g →
$$
\frac{50}{22.7} \times 100\% \approx 220\%
$$
But this is impossible.
→ Error in problem: Actual yield cannot exceed theoretical yield.
→ Likely typo: Perhaps actual yield is 5.0 g or 20 g.
If actual = 20 g, then percent yield ≈ 88.1%.

---

⚠️ Recommendation:

Please verify the value in part 6. It should be less than or equal to 22.7 g. If it’s 50 g, it’s unphysical.