Limiting Reactant and Percent Yield Practice - Free Printable

Final Answer:
The limiting reactant is Na₃PO₄. The mass of (NH₄)₃PO₄ formed is 40.8 grams. The mass of NaNO₃ formed is 37.6 grams. The mass of excess NH₄NO₃ left over is 15.2 grams.

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

Explanation:
Let’s break this down step by step like we’re cooking a recipe.

Step 1: Balance the reaction
The reaction is already balanced:
NH₄NO₃ + Na₃PO₄ → (NH₄)₃PO₄ + NaNO₃
But wait — it’s not actually balanced! Let’s fix it first.
Left side: N=2, H=4, O=6, Na=3, P=1
Right side: N=4, H=12, O=7, Na=1, P=1
So we need to balance it properly.
Correct balanced equation:
3 NH₄NO₃ + Na₃PO₄ → (NH₄)₃PO₄ + 3 NaNO₃

Now it’s balanced! 3 ammonium nitrates + 1 sodium phosphate → 1 ammonium phosphate + 3 sodium nitrates.

Step 2: Find molar masses
- NH₄NO₃ = 14 + 4×1 + 14 + 3×16 = 80 g/mol
- Na₃PO₄ = 3×23 + 31 + 4×16 = 164 g/mol
- (NH₄)₃PO₄ = 3×18 + 31 + 64 = 149 g/mol
- NaNO₃ = 23 + 14 + 48 = 85 g/mol

Step 3: Convert grams to moles
We have 30.0 g NH₄NO₃ → 30.0 / 80 = 0.375 moles
We have 50.0 g Na₃PO₄ → 50.0 / 164 ≈ 0.305 moles

Step 4: Use the balanced equation to find which runs out first
From the equation: 3 moles NH₄NO₃ need 1 mole Na₃PO₄
So for 0.375 moles NH₄NO₃, we need 0.375 ÷ 3 = 0.125 moles Na₃PO₄
But we have 0.305 moles Na₃PO₄ — that’s more than enough. So NH₄NO₃ will run out first? Wait — let’s check the other way.

What if we use all 0.305 moles Na₃PO₄?
It needs 3 × 0.305 = 0.915 moles NH₄NO₃
But we only have 0.375 moles NH₄NO₃ — not enough!
So NH₄NO₃ is limiting? Wait — no, let’s double-check.

Actually, from the ratio:
3 mol NH₄NO₃ : 1 mol Na₃PO₄
So the “mole ratio needed” is 3:1
We have 0.375 mol NH₄NO₃ and 0.305 mol Na₃PO₄
Divide each by their coefficient:
NH₄NO₃: 0.375 ÷ 3 = 0.125
Na₃PO₄: 0.305 ÷ 1 = 0.305
The smaller number (0.125) tells us NH₄NO₃ is limiting? But wait — that can’t be right because we have less Na₃PO₄ in moles.

Wait — I think I made a mistake earlier. Let’s recalculate:

We have:
- NH₄NO₃: 30.0 g / 80 g/mol = 0.375 mol
- Na₃PO₄: 50.0 g / 164 g/mol ≈ 0.305 mol

From the balanced equation: 3 NH₄NO₃ + 1 Na₃PO₄ → products

So to react with 0.305 mol Na₃PO₄, we need 3 × 0.305 = 0.915 mol NH₄NO₃ — but we only have 0.375 mol. Not enough.

To react with 0.375 mol NH₄NO₃, we need 0.375 ÷ 3 = 0.125 mol Na₃PO₄ — and we have 0.305 mol, which is plenty.

So NH₄NO₃ is the limiting reactant? But that contradicts the initial thought.

Wait — let’s think again. If we have 0.375 mol NH₄NO₃, it can only react with 0.125 mol Na₃PO₄. We have 0.305 mol Na₃PO₄, so yes, NH₄NO₃ is limiting.

But the problem says “which reactant is limiting”, and based on calculation, NH₄NO₃ is limiting.

But let me check the answer again. Maybe I misread.

Actually, let’s calculate how much product each can make.

From NH₄NO₃ (limiting):
0.375 mol NH₄NO₃ → since 3 mol NH₄NO₃ make 1 mol (NH₄)₃PO₄, so 0.375 ÷ 3 = 0.125 mol (NH₄)₃PO₄
Mass = 0.125 × 149 = 18.625 g

And 0.375 mol NH₄NO₃ → 3 mol NH₄NO₃ make 3 mol NaNO₃, so 0.375 mol NaNO₃
Mass = 0.375 × 85 = 31.875 g

Now from Na₃PO₄:
0.305 mol Na₃PO₄ → makes 0.305 mol (NH₄)₃PO₄ → 0.305 × 149 = 45.445 g
And 0.305 mol Na₃PO₄ → makes 3 × 0.305 = 0.915 mol NaNO₃ → 0.915 × 85 = 77.775 g

But we don’t have enough NH₄NO₃ to make that much, so NH₄NO₃ is limiting.

But the initial answer I gave said Na₃PO₄ is limiting — that was wrong.

Let me correct this.

Actually, I think I found the error. In the balanced equation, it's 3 NH₄NO₃ + 1 Na₃PO₄ → 1 (NH₄)₃PO₄ + 3 NaNO₃

So the mole ratio is 3:1.

We have 0.375 mol NH₄NO₃ and 0.305 mol Na₃PO₄.

The amount of Na₃PO₄ needed for 0.375 mol NH₄NO₃ is 0.375 / 3 = 0.125 mol.

We have 0.305 mol Na₃PO₄, which is more than 0.125 mol, so Na₃PO₄ is in excess, and NH₄NO₃ is limiting.

Yes, that's correct.

So limiting reactant is NH₄NO₃.

Then products:
- (NH₄)₃PO₄: 0.125 mol × 149 g/mol = 18.6 g
- NaNO₃: 0.375 mol × 85 g/mol = 31.9 g

Excess reactant left: Na₃PO₄ used = 0.125 mol × 164 g/mol = 20.5 g
Started with 50.0 g, so left = 50.0 - 20.5 = 29.5 g

But this contradicts the initial answer I provided.

I think there might be a mistake in the problem or my understanding.

Wait — perhaps the reaction is written as is, and we should use it as given, even if not balanced? But that doesn't make sense.

Looking back at the image, the reaction is written as:
NH₄NO₃ + Na₃PO₄ → (NH₄)₃PO₄ + NaNO₃

This is not balanced. So we must balance it first.

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

So with that, as above, NH₄NO₃ is limiting.

But let's calculate again.

Moles:
NH₄NO₃: 30.0 / 80 = 0.375 mol
Na₃PO₄: 50.0 / 164 = 0.3049 mol ≈ 0.305 mol

From equation, 3 mol NH₄NO₃ require 1 mol Na₃PO₄.
So 0.375 mol NH₄NO₃ require 0.375/3 = 0.125 mol Na₃PO₄.
Available Na₃PO₄ is 0.305 mol > 0.125 mol, so NH₄NO₃ is limiting.

Products:
(NH₄)₃PO₄: from 3 mol NH₄NO₃ -> 1 mol (NH₄)₃PO₄, so 0.375 mol NH₄NO₃ -> 0.125 mol (NH₄)₃PO₄
Mass = 0.125 * 149 = 18.625 g ≈ 18.6 g

NaNO₃: 3 mol NH₄NO₃ -> 3 mol NaNO₃, so 0.375 mol NH₄NO₃ -> 0.375 mol NaNO₃
Mass = 0.375 * 85 = 31.875 g ≈ 31.9 g

Excess Na₃PO₄ used: 0.125 mol * 164 g/mol = 20.5 g
Left over: 50.0 - 20.5 = 29.5 g

So the correct answer should be:
Limiting reactant: NH₄NO₃
Mass of (NH₄)₃PO₄: 18.6 g
Mass of NaNO₃: 31.9 g
Excess Na₃PO₄ left: 29.5 g

But this is different from what I initially said.

Perhaps the user expects the reaction as written, unbalanced, but that would be incorrect.

Maybe I misread the initial answer.

Given the time, I think there was a mistake in the first response.

For the sake of this, I'll provide the correct calculation.

Final Answer:
The limiting reactant is NH₄NO₃. The mass of (NH₄)₃PO₄ formed is 18.6 grams. The mass of NaNO₃ formed is 31.9 grams. The mass of excess Na₃PO₄ left over is 29.5 grams.