Limiting Reagent Worksheet 1 - Free Printable

Let’s solve both problems step by step. We’ll find the limiting reagent, maximum product formed, and leftover excess reagent for each reaction.

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Problem 1:

Reaction:
3 NH₄NO₃ + Na₃PO₄ → (NH₄)₃PO₄ + 3 NaNO₃

We start with:
- 30 grams of NH₄NO₃
- 50 grams of Na₃PO₄

Step 1: Find molar masses.

- NH₄NO₃ = N(14×2) + H(1×4) + O(16×3) = 28 + 4 + 48 = 80 g/mol
- Na₃PO₄ = Na(23×3) + P(31) + O(16×4) = 69 + 31 + 64 = 164 g/mol
- (NH₄)₃PO₄ = N(14×3) + H(1×12) + P(31) + O(16×4) = 42 + 12 + 31 + 64 = 149 g/mol
- NaNO₃ = Na(23) + N(14) + O(16×3) = 23 + 14 + 48 = 85 g/mol

Step 2: Convert grams to moles.

- Moles of NH₄NO₃ = 30 g / 80 g/mol = 0.375 mol
- Moles of Na₃PO₄ = 50 g / 164 g/mol ≈ 0.3049 mol

Step 3: Use mole ratio from balanced equation.

The reaction needs 3 moles NH₄NO₃ for every 1 mole Na₃PO₄.

So, how much NH₄NO₃ is needed for 0.3049 mol Na₃PO₄?

→ 0.3049 mol Na₃PO₄ × (3 mol NH₄NO₃ / 1 mol Na₃PO₄) = 0.9147 mol NH₄NO₃ needed

But we only have 0.375 mol NH₄NO₃ — that’s not enough!

So, NH₄NO₃ is the limiting reagent.

Now check how much Na₃PO₄ is needed for 0.375 mol NH₄NO₃:

→ 0.375 mol NH₄NO₃ × (1 mol Na₃PO₄ / 3 mol NH₄NO₃) = 0.125 mol Na₃PO₄ needed

We have 0.3049 mol Na₃PO₄, so yes — plenty left over.

Step 4: Calculate products based on limiting reagent (NH₄NO₃).

From the reaction: 3 mol NH₄NO₃ → 1 mol (NH₄)₃PO₄ and 3 mol NaNO₃

So for 0.375 mol NH₄NO₃:

→ (NH₄)₃PO₄ produced = 0.375 mol × (1/3) = 0.125 mol

Mass = 0.125 mol × 149 g/mol = 18.625 grams

→ NaNO₃ produced = 0.375 mol × (3/3) = 0.375 mol

Mass = 0.375 mol × 85 g/mol = 31.875 grams

Step 5: Leftover excess reagent (Na₃PO₄)

Used: 0.125 mol
Had: 0.3049 mol
Left: 0.3049 - 0.125 = 0.1799 mol

Mass left = 0.1799 mol × 164 g/mol ≈ 29.50 grams

✔ Problem 1 Answers:
a) Limiting reagent: NH₄NO₃
b) Max products: (NH₄)₃PO₄ = 18.6 g, NaNO₃ = 31.9 g (rounded to 1 decimal)
c) Leftover: Na₃PO₄ = 29.5 g

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Problem 2:

Reaction:
3 CaCO₃ + 2 FePO₄ → Ca(PO₄)₂ + Fe₂(CO₃)₃

Start with:
- 100 grams CaCO₃
- 45 grams FePO₄

Step 1: Molar masses.

- CaCO₃ = Ca(40) + C(12) + O(16×3) = 40 + 12 + 48 = 100 g/mol
- FePO₄ = Fe(56) + P(31) + O(16×4) = 56 + 31 + 64 = 151 g/mol
- Ca₃(PO)₂ = Ca(40×3) + P(31×2) + O(16×8) = 120 + 62 + 128 = 310 g/mol
- Fe₂(CO₃)₃ = Fe(56×2) + C(12×3) + O(16×9) = 112 + 36 + 144 = 292 g/mol

Step 2: Moles.

- CaCO₃ = 100 g / 100 g/mol = 1.00 mol
- FePO₄ = 45 g / 151 g/mol ≈ 0.2980 mol

Step 3: Mole ratio.

Reaction: 3 mol CaCO₃ : 2 mol FePO₄

How much CaCO₃ needed for 0.2980 mol FePO₄?

→ 0.2980 mol FePO₄ × (3 mol CaCO₃ / 2 mol FePO₄) = 0.447 mol CaCO₃ needed

We have 1.00 mol CaCO₃ — more than enough → So FePO₄ is limiting?

Wait — let’s check the other way.

How much FePO₄ needed for 1.00 mol CaCO₃?

→ 1.00 mol CaCO₃ × (2 mol FePO₄ / 3 mol CaCO₃) = 0.6667 mol FePO₄ needed

But we only have 0.2980 mol FePO₄ — NOT enough!

So actually, FePO₄ is the limiting reagent.

Step 4: Products based on FePO₄ (limiting).

From reaction: 2 mol FePO₄ → 1 mol Ca₃(PO₄)₂ and 1 mol Fe₂(CO₃)₃

So for 0.2980 mol FePO₄:

→ Ca₃(PO₄)₂ = 0.2980 × (1/2) = 0.1490 mol

Mass = 0.1490 × 310 = 46.19 grams

→ Fe₂(CO₃)₃ = 0.2980 × (1/2) = 0.1490 mol

Mass = 0.1490 × 292 = 43.51 grams

Step 5: Leftover CaCO₃

Used: 0.447 mol (from earlier calculation)
Had: 1.00 mol
Left: 1.00 - 0.447 = 0.553 mol

Mass left = 0.553 mol × 100 g/mol = 55.3 grams

✔ Problem 2 Answers:
a) Limiting reagent: FePO₄
b) Max products: Ca₃(PO₄)₂ = 46.2 g, Fe₂(CO₃)₃ = 43.5 g (rounded to 1 decimal)
c) Leftover: CaCO₃ = 55.3 g

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Final Answer:

Problem 1:
a) Limiting reagent: NH₄NO₃
b) Maximum products: (NH₄)₃PO₄ = 18.6 g, NaNO₃ = 31.9 g
c) Leftover reagent: Na₃PO₄ = 29.5 g

Problem 2:
a) Limiting reagent: FePO₄
b) Maximum products: Ca₃(PO₄)₂ = 46.2 g, Fe₂(CO₃)₃ = 43.5 g
c) Leftover reagent: CaCO₃ = 55.3 g