Quiz & Worksheet - How to Calculate Percent Yield

Final Answer:
38%

──────────────────────────────────────

Explanation:
Let’s break this down step by step — it’s like a recipe!

We’re given:
- 6.5 grams of C₆H₁₂O₆ (glucose) reacts.
- It makes 2.5 grams of CO₂ (carbon dioxide) in real life (this is the “actual yield”).
- We need to find the percent yield — which means: “How good was the reaction compared to what it should’ve made?”

Step 1: Find out how much CO₂ we *should* get if everything went perfectly.

Look at the reaction:
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

That means 1 molecule of glucose makes 2 molecules of CO₂.

Now, we need to convert grams to moles because chemistry works with molecules, not grams directly.

Molar mass of glucose (C₆H₁₂O₆):
- C: 6 × 12 = 72
- H: 12 × 1 = 12
- O: 6 × 16 = 96
Total = 72 + 12 + 96 = 180 g/mol

So, 6.5 grams of glucose = 6.5 ÷ 180 = 0.0361 moles

From the reaction: 1 mole glucose → 2 moles CO₂
So, 0.0361 moles glucose → 0.0722 moles CO₂

Now, molar mass of CO₂:
- C: 12, O: 16×2 = 32 → Total = 44 g/mol

So, 0.0722 moles CO₂ × 44 g/mol = 3.177 grams

That’s the “theoretical yield” — what we should get if the reaction was perfect.

Step 2: Percent Yield = (Actual Yield ÷ Theoretical Yield) × 100

Actual = 2.5 grams
Theoretical = 3.177 grams

→ (2.5 ÷ 3.177) × 100 ≈ 78.7% → Wait! That’s not one of the choices?

Hold on — let me double-check.

Wait — maybe I misread the reaction? Let me check again.

Actually, looking back — the question says “produces 2.5 grams of CO₂”, and options are 120%, 79%, 39%, 38%.

My math gave me ~79%. But that’s not matching the answer I wrote above.

Wait — did I make a mistake?

Let me recalculate:

6.5 g glucose ÷ 180 g/mol = 0.03611 mol

→ Makes 2 × 0.03611 = 0.07222 mol CO₂

→ 0.07222 mol × 44 g/mol = 3.1777 g CO₂ (theoretical)

Actual = 2.5 g

Percent yield = (2.5 / 3.1777) × 100 = 78.67% → rounds to 79%

But the correct answer listed is 38%? That doesn’t match.

Wait — maybe the question is different? Or maybe I misread the formula?

Looking again — the reaction is: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ — that’s correct.

Unless... maybe the 2.5 grams is not CO₂? No, it says CO₂.

Wait — perhaps there’s a trick?

Another thought — maybe they want the yield based on something else?

No — percent yield is always (actual / theoretical) × 100.

Hmm… unless I made a calculation error?

Let me try using exact numbers:

6.5 / 180 = 65/1800 = 13/360 ≈ 0.036111

× 2 = 0.072222 mol CO₂

× 44 = 3.17777... g

2.5 / 3.17777 = 0.7867 → 78.67% → 79%

So why is the answer 38%?

Wait — maybe the question is asking for something else? Or maybe it’s a typo?

Actually — looking at the image again — I think I see the issue.

The user uploaded an image with multiple questions. The first question is about glucose making CO₂, and the correct answer according to the quiz is likely 38% — but my math says 79%.

This suggests either:

1. There’s a mistake in the problem or options.
2. I misread the reaction.

Wait — let me check the reaction again.

C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ — that’s fermentation. Correct.

Maybe the 2.5 grams is not of CO₂? No, it says CO₂.

Perhaps the question is asking for percent yield of ethanol? But no, it says CO₂.

Alternatively — maybe the molar mass is different?

C₆H₁₂O₆: 6×12=72, 12×1=12, 6×16=96 → 180 — correct.

CO₂: 12+32=44 — correct.

6.5g glucose → ? g CO₂

Ratio: 180g glucose → 88g CO₂ (because 2×44=88)

So, 6.5g glucose → (88/180) × 6.5 = ?

Calculate: 88 ÷ 180 = 0.48888...

× 6.5 = 3.1777... g — same as before.

Actual = 2.5g

2.5 / 3.1777 = 0.7867 → 78.67% → 79%

But the options include 79% — so why did I write 38% earlier?

Ah — I think I confused myself. In my initial response, I said “Final Answer: 38%” — but that’s wrong.

The correct answer should be 79%.

But let me check the other questions quickly.

Question 2: 125g C₆H₆O₃ → 51g H₂O

Reaction: C₆H₆O₃ + 3O₂ → 6CO + 3H₂O

Molar mass C₆H₆O₃: 6×12=72, 6×1=6, 3×16=48 → 126 g/mol

125g ÷ 126 = 0.992 mol

Makes 3 mol H₂O per 1 mol C₆H₆O₃ → 2.976 mol H₂O

Molar mass H₂O = 18 → 2.976 × 18 = 53.57 g theoretical

Actual = 51g

Percent yield = 51 / 53.57 × 100 ≈ 95.2% → 95% — matches option.

Question 3: 5.00g KClO₃ → 1.5g O₂

Reaction: 2KClO₃ → 2KCl + 3O₂

Molar mass KClO₃: 39+35.5+48=122.5 g/mol

5.00g ÷ 122.5 = 0.04082 mol

From reaction: 2 mol KClO₃ → 3 mol O₂

So 0.04082 mol KClO₃ → (3/2)×0.04082 = 0.06123 mol O₂

Molar mass O₂ = 32 → 0.06123 × 32 = 1.959 g theoretical

Actual = 1.5g

Percent yield = 1.5 / 1.959 × 100 ≈ 76.57% → 77% — matches option.

So for Question 1, my calculation gives 79%, and it’s an option.

Therefore, the correct answer is 79%.

I must have typed 38% by mistake earlier.

Let me correct that.

Final Answer:
79%