Limiting Reactants Worksheet | GCSE Chemistry | Beyond - Free Printable

Let’s solve this step by step.

We’re given the chemical reaction:

NaHCO₃ + CHCOOH → CH₃COONa + CO₂ + H₂O

This is already balanced — 1 molecule of each reactant makes 1 molecule of each product. So we don’t need to adjust coefficients.

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Step 1: Calculate molar masses

We’ll use atomic masses (rounded for simplicity):

- Na = 23 g/mol
- H = 1 g/mol
- C = 12 g/mol
- O = 16 g/mol
- For CH₃COOH, remember it has 2 C, 4 H, and 2 O atoms.

#### Molar mass of NaHCO₃:
Na + H + C + 3×O = 23 + 1 + 12 + 48 = 84 g/mol

#### Molar mass of CH₃COOH:
2×C + 4×H + 2×O = 24 + 4 + 32 = 60 g/mol

#### Molar mass of CO₂:
C + 2×O = 12 + 32 = 44 g/mol

*(We don’t need others for these questions, but good to know they’re there.)*

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Step 2: Question 2 – Which reactant is limiting?

We are told:
→ 5g of sodium bicarbonate (NaHCO₃)
→ 9g of acetic acid (CH₃COOH)

From the balanced equation:
1 mole NaHCO₃ reacts with 1 mole CH₃COOH

So let’s find how many moles we have of each.

#### Moles of NaHCO₃:
mass / molar mass = 5g / 84 g/mol ≈ 0.0595 mol

#### Moles of CH₃COOH:
9g / 60 g/mol = 0.15 mol

Since the reaction needs them in a 1:1 ratio, and we have less NaHCO₃ (0.0595 mol) than CH₃COOH (0.15 mol), NaHCO₃ will run out first.

✔ So, sodium bicarbonate (NaHCO₃) is the limiting reactant.

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Step 3: Question 4 – What mass of carbon dioxide do you expect?

The limiting reactant determines how much product forms.

From the equation:
1 mole NaHCO₃ → 1 mole CO₂

So 0.0595 mol NaHCO₃ → 0.0595 mol CO₂

Now convert moles of CO₂ to grams:

mass = moles × molar mass = 0.0595 mol × 44 g/mol ≈ ?

Let’s calculate:

0.0595 × 44 =
First, 0.06 × 44 = 2.64
But 0.0595 is 0.0005 less than 0.06 → 0.0005 × 44 = 0.022
So 2.64 - 0.022 = 2.618 g

Or directly:
0.0595 × 44 = (595/10000) × 44 = (595 × 44) / 10000
595 × 44 = 595 × 40 + 595 × 4 = 23800 + 2380 = 26180
26180 / 10000 = 2.618 g

Rounded reasonably: ≈ 2.62 g

*(Note: If your teacher wants exact fractions, we can do that too — but decimal is fine here.)*

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Step 4: Question 5 – Why might actual yield differ?

Even if we calculated 2.62 g, the student only got 2.3 g.

Possible reasons:

- Not all the reaction happened (maybe not mixed well).
- Some gas escaped before being measured.
- Impurities in the chemicals.
- Measurement errors (scale not accurate, etc.).
- Reaction didn’t go to completion.

Any of these could cause less product than expected.

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Final Answer:

Limiting reactant: Sodium bicarbonate (NaHCO₃)
Expected mass of CO₂: 2.62 g
Reason for lower yield: Possible experimental error like incomplete reaction or loss of gas.