Solved Name: Stoichiometry Percent Yield Worksheet SHOW ALL ... - Free Printable

Let’s solve this step by step.

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We are given the unbalanced chemical equation:

FePO₄ + Na₂SO₄ → Fe₂(SO₄)₃ + Na₃PO₄

Step 1: Balance the equation

Look at each element:

- Iron (Fe): Left has 1, right has 2 → put 2 in front of FePO₄
- Phosphate (PO₄): Now left has 2 PO₄, right has 1 → put 2 in front of Na₃PO₄
- Sodium (Na): Right now has 6 Na (from 2 × Na₃), so left needs 3 Na₂SO₄ (since 3 × 2 = 6)
- Sulfate (SO₄): Left has 3 SO₄ (from 3 Na₂SO₄), right has 3 SO₄ (in Fe₂(SO₄)₃) → good!

Balanced equation:

2 FePO₄ + 3 Na₂SO₄ → Fe₂(SO₄)₃ + 2 Na₃PO₄

✔ Balanced!

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## Part a) Theoretical yield of iron(III) sulfate from 25 g FePO₄ (with excess Na₂SO₄)

We need to find how many grams of Fe₂(SO₄)₃ we can make from 25 g of FePO₄.

This is a stoichiometry problem.

Step 1: Molar masses

- FePO₄:
Fe = 55.85, P = 30.97, O = 16.00 × 4 = 64.00
→ 55.85 + 30.97 + 64.00 = 150.82 g/mol

- Fe₂(SO₄)₃:
Fe = 55.85 × 2 = 111.70
S = 32.07 × 3 = 96.21
O = 16.00 × 12 = 192.00
→ 111.70 + 96.21 + 192.00 = 399.91 g/mol

Step 2: Moles of FePO₄ used

Moles = mass / molar mass = 25 g / 150.82 g/mol ≈ 0.1658 mol

Step 3: Use mole ratio from balanced equation

From balanced equation:
2 mol FePO₄ → 1 mol Fe₂(SO₄)₃

So moles of Fe₂(SO₄)₃ = 0.1658 mol FePO₄ × (1 mol Fe₂(SO₄)₃ / 2 mol FePO₄) = 0.0829 mol

Step 4: Convert to grams

Mass = moles × molar mass = 0.0829 mol × 399.91 g/mol ≈ 33.15 g

✔ So theoretical yield = 33.15 grams of Fe₂(SO₄)₃

*(We’ll round to 2 decimal places for now — but keep more digits if needed later.)*

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## Part b) Percent yield if actual yield is 18.5 g

Formula:

% Yield = (Actual Yield / Theoretical Yield) × 100

= (18.5 / 33.15) × 100 ≈ ?

Calculate: 18.5 ÷ 33.15 ≈ 0.5581 → × 100 = 55.81%

✔ Percent yield ≈ 55.8%

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## Part c) Is the answer from problem #3 reasonable? Explain.

Wait — there is no “problem #3” shown on the worksheet. But looking at the structure, part (c) probably refers to part (b)’s percent yield (55.8%).

Is 55.8% reasonable?

Yes! In real lab experiments, you rarely get 100% yield because of spills, incomplete reactions, side reactions, or loss during purification. A yield between 50–80% is common and reasonable. 55.8% is a bit low but still plausible — maybe some product was lost or reaction didn’t go to completion.

✔ Answer: Yes, it’s reasonable. Real-world reactions often have less than 100% yield due to experimental losses or inefficiencies.

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## Part d) If we use 15 g sodium sulfate and get 65.0% yield, how many grams of sodium phosphate will I make?

Now we’re starting with Na₂SO₄, not FePO₄. And we want Na₃PO₄ as product.

Also, note: We’re told the yield is 65.0%, so we must calculate theoretical yield first, then multiply by 0.65.

Balanced equation again:

2 FePO₄ + 3 Na₂SO₄ → Fe₂(SO₄)₃ + 2 Na₃PO₄

We’re using 15 g Na₂SO₄. Assume FePO₄ is in excess? The problem doesn’t say, but since it says “if I do this reaction with 15 grams of sodium sulfate”, and doesn’t mention limiting reactant, we assume Na₂SO₄ is the limiting reactant.

Step 1: Molar mass of Na₂SO₄

Na = 22.99 × 2 = 45.98
S = 32.07
O = 16.00 × 4 = 64.00
→ Total = 45.98 + 32.07 + 64.00 = 142.05 g/mol

Step 2: Moles of Na₂SO₄

Moles = 15 g / 142.05 g/mol ≈ 0.1056 mol

Step 3: Mole ratio to Na₃PO₄

From balanced equation:
3 mol Na₂SO₄ → 2 mol Na₃PO₄

So moles of Na₃PO₄ = 0.1056 mol Na₂SO₄ × (2 mol Na₃PO₄ / 3 mol Na₂SO₄) = 0.0704 mol

Step 4: Molar mass of Na₃PO₄

Na = 22.99 × 3 = 68.97
P = 30.97
O = 16.00 × 4 = 64.00
→ Total = 68.97 + 30.97 + 64.00 = 163.94 g/mol

Step 5: Theoretical yield of Na₃PO₄

Mass = 0.0704 mol × 163.94 g/mol ≈ 11.54 g

Step 6: Apply 65.0% yield

Actual yield = 65.0% of theoretical = 0.650 × 11.54 g ≈ 7.50 g

✔ Final answer for part d: 7.50 grams of sodium phosphate

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Final Answer:
a) 33.15 grams
b) 55.8%
c) Yes, it's reasonable because real reactions often have less than 100% yield due to experimental losses.
d) 7.50 grams