How to identify the limiting reactant?

Question

I'm studying chemical reactions involving multiple reactants, and I'm having trouble determining the limiting reactant. For example, if there's a mixture of H₂SO₄ and NaOH reacting together in an unknown ratio, how can I identify the limiting reactant if the problem only provides the mass or the concentration and volume? Is there a general method to solve this quickly and accurately?

Thank you!

(Source: space waves)

Answer 1

1) Only mass given

• Convert to moles: moles = grams ÷ molar mass

• Use the balanced equation: moles ÷ coefficient (smaller result = limiting reactant)

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

given 49g H₂SO₄ and 10g NaOH

convert to moles (moles = grams provided/molar mass from periodic table)

49g H₂SO₄ ÷ 98 g/mol = 0.50 mol H₂SO₄

10g NaOH ÷ 40 g/mol = 0.25 mol NaOH

Use answer from above and coefficients from balanced equation

0.5 mol H₂SO₄ ÷ 1 = 0.50 mol

0.25 mol NaOH ÷ 2 = 0.125 mol (limiting reactant is smallest number)

2) Only concentration & volume given

• Find moles: moles = M × V (in liters)

• Use the balanced equation: moles ÷ coefficient (smaller result = limiting reactant)

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

given 0.50L of 2.0M H₂SO₄ and 0.60L of 1.5M NaOH

convert to moles (moles = grams provided/molar mass from periodic table)

2.0 mol/L x 0.50 L = 1.00 mol H₂SO₄

1.5 mol/L x 0.60 L = 0.90 mol NaOH

use answer from above and coefficients from balanced equation

1.00 ÷ 1 = 0.5 mol H₂SO₄

0.90 ÷ 2 = 0.45 mol NaOH (limiting reactant)

Answer 2

How to Identify the Limiting Reactant (General Method):

1. Write the balanced chemical equation.
2. Example: H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O
3. Convert given amounts to moles.
4. If given mass: moles= mass (g) ÷ molar mass (g/mol)
5. If given solution: moles= volume (L) × concentration (mol/L)
6. Calculate the “required” ratio for each reactant.
7. Divide the moles you have by the coefficient from the balanced equation:

Formula:

Limiting Reactant value = (moles available) / (moles required from coefficients)

1. Identify the limiting reactant.
2. The reactant with the smallest Limiting Reactant value is the limiting reactant.

Example:

1. Balanced equation: H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O
2. Given: 50 g H₂SO₄ (M = 98 g/mol), 40 g NaOH (M = 40 g/mol)

Step 1: Convert to moles

1. H₂SO₄: 50 ÷ 98 ≈ 0.51 mol
2. NaOH: 40 ÷ 40 = 1.00 mol

Step 2: Divide by coefficients

1. H₂SO₄: 0.51 ÷ 1 = 0.51
2. NaOH: 1.00 ÷ 2 = 0.50

Step 3: Compare

1. NaOH has smaller value → NaOH is limiting
2. H₂SO₄ is in excess

Answer 3

Write the balanced equation:

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O\text{H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O}H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

Convert given amounts to moles:

Suppose you have 50 g of H₂SO₄ and 40 g of NaOH.

1. Moles of H₂SO₄ = 50 ÷ 98 ≈ 0.51 mol
2. Moles of NaOH = 40 ÷ 40 ≈ 1.00 mol

Compare mole ratio with the balanced equation:

The equation needs 2 moles of NaOH for every 1 mole of H₂SO₄.

1. Available ratio: 1.00 ÷ 0.51 ≈ 1.96 ≈ 2 → just enough NaOH, but slightly less than needed.

Identify limiting reactant:

H₂SO₄ would require 1.02 mol NaOH to react completely (0.51 × 2), but only 1.00 mol is available.

1. NaOH is the limiting reactant because it will run out first.

Conclusion:

The reaction stops when NaOH is used up, leaving some H₂SO₄ unreacted.

Answer 4

One easy and quick way to find the limiting reactant is to simply divide the moles of each reactant by the corresponding coefficient in the balanced equation, and whichever result is the lower value indicates the limiting reactant. Hopefully you already know how to find moles given grams or given volume and molarity.

in your example, the balanced equation is

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

If, for example, you have 1 mole of H2SO4 and 1 mole of NaOH, dividing each by their coefficient, you get a value of 1 for H2SO4 and a value of 0.5 for NaOH. So, NaOH is the limiting reactant.

Answer 5

First you write the balance equation: H2SO4 + 2NaOH -> Na2SO4 + 2H2O}

this indicates that 1 mole of H₂SO₄ reacts with 2 moles of NaOH.

Then convert given quantities to moles, for example if you were given 98 grams of H₂SO₄ (molar mass ≈ 98 g/mol), so 98g/98g/m=1mol

so in this case since it is 40g/mol for NaOH molar mass, and you need 2 moles of NaOH with each H₂SO₄, then to neutralize the reaction takes 80g of NaOH and anything less would make NaOH the limiting reagent.

if given volume and concentration the moles=Volume (L) × Concentration (mol/L)

reactant with the smallest resulting value is the limiting reagent