How to Solve: Limiting Reagent

How to Solve: Limiting Reagent

Introduction

"You’re baking cookies, but you run out of chocolate chips—now the whole batch is ruined. In chemistry, the ‘chocolate chips’ are your limiting reagent, and if you don’t find it first, your whole reaction fails. Master this, and you’ll nail every stoichiometry problem on your exam."

What You Need To Know First

Before tackling limiting reagents, you must understand: 1. Balanced chemical equations – Coefficients tell you the mole ratio of reactants and products. 2. Mole calculations – How to convert between mass, moles, and particles (Avogadro’s number). 3. Stoichiometry – Using mole ratios to predict product amounts from reactant amounts.

If any of these are shaky, stop here and review them first.

Key Vocabulary

Formulas To Know

1. Moles from mass
   [
   n = \frac{m}{M}
   ]
2. ( n ) = number of moles (mol)
3. ( m ) = mass (g)
4. ( M ) = molar mass (g/mol)

5. MEMORISE THIS – You’ll use it in every problem.

6. Mole ratio from balanced equation
   [
   \text{Mole ratio} = \frac{\text{Coefficient of desired substance}}{\text{Coefficient of given substance}}
   ]

7. Given on exam sheet (but you must know how to apply it).

8. Theoretical yield (mass of product)
   [
   m_{\text{product}} = n_{\text{limiting reagent}} \times \text{Mole ratio} \times M_{\text{product}}
   ]

9. MEMORISE THIS – This is how you calculate the maximum product possible.

Step-by-Step Method

Follow these steps exactly for every limiting reagent problem.

1. Write the balanced chemical equation.

2. If it’s not given, balance it yourself.

3. Convert all given reactant masses to moles.

4. Use ( n = \frac{m}{M} ).

5. Pick one reactant and calculate how many moles of the other reactant are needed to fully react with it.

6. Use the mole ratio from the balanced equation.

7. Compare the calculated moles needed to the actual moles available.

8. If needed > available, the other reactant is limiting.

9. If needed < available, the reactant you picked is limiting.

10. Use the limiting reagent to calculate the theoretical yield of the product.

11. Multiply moles of limiting reagent by the mole ratio, then by the product’s molar mass.

12. If asked, calculate the mass of excess reagent left over.

13. Subtract the moles used from the moles available, then convert to mass.

Worked Example (Using the Steps Above)

Problem: 2.50 g of hydrogen gas (H₂) reacts with 10.0 g of oxygen gas (O₂) to form water (H₂O). a) Identify the limiting reagent. b) Calculate the theoretical yield of water (in grams).

Solution:

1. Balanced equation:
   [
   2H₂ + O₂ → 2H₂O
   ]

2. Convert masses to moles:

3. Molar mass of H₂ = 2.02 g/mol
   [
   n_{H₂} = \frac{2.50 \text{ g}}{2.02 \text{ g/mol}} = 1.24 \text{ mol}
   ]

4. Molar mass of O₂ = 32.00 g/mol
   [
   n_{O₂} = \frac{10.0 \text{ g}}{32.00 \text{ g/mol}} = 0.313 \text{ mol}
   ]

5. Calculate moles of O₂ needed to react with 1.24 mol H₂:

6. Mole ratio (H₂:O₂) = 2:1
   [
   n_{O₂ \text{ needed}} = \frac{1.24 \text{ mol H₂}}{2} = 0.620 \text{ mol O₂}
   ]

7. Compare needed vs. available O₂:

8. Needed: 0.620 mol
9. Available: 0.313 mol

10. 0.620 > 0.313 → O₂ is limiting (we don’t have enough O₂ to use all the H₂).

11. Calculate theoretical yield of H₂O:

12. Mole ratio (O₂:H₂O) = 1:2
    [
    n_{H₂O} = 0.313 \text{ mol O₂} \times 2 = 0.626 \text{ mol H₂O}
    ]

13. Molar mass of H₂O = 18.02 g/mol
    [
    m_{H₂O} = 0.626 \text{ mol} \times 18.02 \text{ g/mol} = 11.3 \text{ g}
    ]

14. (Optional) Calculate excess H₂ left over:

15. Moles of H₂ used:
    [
    n_{H₂ \text{ used}} = 0.313 \text{ mol O₂} \times 2 = 0.626 \text{ mol H₂}
    ]
16. Moles of H₂ left:
    [
    1.24 \text{ mol} - 0.626 \text{ mol} = 0.614 \text{ mol}
    ]
17. Mass of H₂ left:
    [
    0.614 \text{ mol} \times 2.02 \text{ g/mol} = 1.24 \text{ g}
    ]

Answer: a) The limiting reagent is O₂. b) The theoretical yield of water is 11.3 g.

Worked Examples

Example 1 – Basic (No Tricks)

Problem: 5.00 g of magnesium (Mg) reacts with 5.00 g of hydrochloric acid (HCl) to form magnesium chloride (MgCl₂) and hydrogen gas (H₂). a) Write the balanced equation. b) Identify the limiting reagent. c) Calculate the mass of MgCl₂ produced.

Solution:

1. Balanced equation:
   [
   Mg + 2HCl → MgCl₂ + H₂
   ]

2. Convert masses to moles:

3. Molar mass of Mg = 24.31 g/mol
   [
   n_{Mg} = \frac{5.00 \text{ g}}{24.31 \text{ g/mol}} = 0.206 \text{ mol}
   ]

4. Molar mass of HCl = 36.46 g/mol
   [
   n_{HCl} = \frac{5.00 \text{ g}}{36.46 \text{ g/mol}} = 0.137 \text{ mol}
   ]

5. Calculate moles of HCl needed to react with 0.206 mol Mg:

6. Mole ratio (Mg:HCl) = 1:2
   [
   n_{HCl \text{ needed}} = 0.206 \text{ mol Mg} \times 2 = 0.412 \text{ mol HCl}
   ]

7. Compare needed vs. available HCl:

8. Needed: 0.412 mol
9. Available: 0.137 mol

10. 0.412 > 0.137 → HCl is limiting.

11. Calculate mass of MgCl₂ produced:

12. Mole ratio (HCl:MgCl₂) = 2:1
    [
    n_{MgCl₂} = \frac{0.137 \text{ mol HCl}}{2} = 0.0685 \text{ mol}
    ]
13. Molar mass of MgCl₂ = 95.21 g/mol
    [
    m_{MgCl₂} = 0.0685 \text{ mol} \times 95.21 \text{ g/mol} = 6.52 \text{ g}
    ]

Answer: a) Balanced equation: Mg + 2HCl → MgCl₂ + H₂ b) Limiting reagent: HCl c) Mass of MgCl₂ produced: 6.52 g

What we did and why: - We converted masses to moles because stoichiometry is based on mole ratios, not masses. - We compared the needed vs. available moles to find the limiting reagent. - We used the limiting reagent to calculate the product because it determines how much reaction can happen.

Example 2 – Medium (Added Complication: Percent Yield)

Problem: 10.0 g of aluminum (Al) reacts with 20.0 g of chlorine gas (Cl₂) to form aluminum chloride (AlCl₃). The actual yield is 15.0 g. Calculate the percent yield.

Solution:

1. Balanced equation:
   [
   2Al + 3Cl₂ → 2AlCl₃
   ]

2. Convert masses to moles:

3. Molar mass of Al = 26.98 g/mol
   [
   n_{Al} = \frac{10.0 \text{ g}}{26.98 \text{ g/mol}} = 0.371 \text{ mol}
   ]

4. Molar mass of Cl₂ = 70.90 g/mol
   [
   n_{Cl₂} = \frac{20.0 \text{ g}}{70.90 \text{ g/mol}} = 0.282 \text{ mol}
   ]

5. Calculate moles of Cl₂ needed to react with 0.371 mol Al:

6. Mole ratio (Al:Cl₂) = 2:3
   [
   n_{Cl₂ \text{ needed}} = 0.371 \text{ mol Al} \times \frac{3}{2} = 0.557 \text{ mol Cl₂}
   ]

7. Compare needed vs. available Cl₂:

8. Needed: 0.557 mol
9. Available: 0.282 mol

10. 0.557 > 0.282 → Cl₂ is limiting.

11. Calculate theoretical yield of AlCl₃:

12. Mole ratio (Cl₂:AlCl₃) = 3:2
    [
    n_{AlCl₃} = 0.282 \text{ mol Cl₂} \times \frac{2}{3} = 0.188 \text{ mol}
    ]

13. Molar mass of AlCl₃ = 133.33 g/mol
    [
    m_{AlCl₃} = 0.188 \text{ mol} \times 133.33 \text{ g/mol} = 25.1 \text{ g}
    ]

14. Calculate percent yield:
    [
    \text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 = \left( \frac{15.0 \text{ g}}{25.1 \text{ g}} \right) \times 100 = 59.8\%
    ]

Answer: Percent yield = 59.8%

What we did and why: - We found the limiting reagent first because it determines the maximum possible product (theoretical yield). - Percent yield compares what we actually got to what we could have gotten—this tells us how efficient the reaction was.

Example 3 – Exam Style (Disguised Problem)

Problem: A student mixes 15.0 mL of 0.500 M sodium hydroxide (NaOH) with 10.0 mL of 0.300 M sulfuric acid (H₂SO₄). The reaction is: [ 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O ] Which reactant is limiting, and what mass of sodium sulfate (Na₂SO₄) is produced?

Solution:

1. Calculate moles of each reactant:
2. Moles of NaOH:
   [
   n_{NaOH} = C \times V = 0.500 \text{ M} \times 0.0150 \text{ L} = 0.00750 \text{ mol}
   ]

3. Moles of H₂SO₄:
   [
   n_{H₂SO₄} = 0.300 \text{ M} \times 0.0100 \text{ L} = 0.00300 \text{ mol}
   ]

4. Calculate moles of H₂SO₄ needed to react with 0.00750 mol NaOH:

5. Mole ratio (NaOH:H₂SO₄) = 2:1
   [
   n_{H₂SO₄ \text{ needed}} = \frac{0.00750 \text{ mol NaOH}}{2} = 0.00375 \text{ mol H₂SO₄}
   ]

6. Compare needed vs. available H₂SO₄:

7. Needed: 0.00375 mol
8. Available: 0.00300 mol

9. 0.00375 > 0.00300 → H₂SO₄ is limiting.

10. Calculate mass of Na₂SO₄ produced:

11. Mole ratio (H₂SO₄:Na₂SO₄) = 1:1
    [
    n_{Na₂SO₄} = 0.00300 \text{ mol}
    ]
12. Molar mass of Na₂SO₄ = 142.04 g/mol
    [
    m_{Na₂SO₄} = 0.00300 \text{ mol} \times 142.04 \text{ g/mol} = 0.426 \text{ g}
    ]

Answer: Limiting reagent: H₂SO₄ Mass of Na₂SO₄ produced: 0.426 g

What we did and why: - We used molarity × volume to find moles because the reactants were given in solution. - The limiting reagent was not the one with fewer moles—it was the one that ran out first based on the mole ratio. - Always check the balanced equation to get the correct mole ratio.

Common Mistakes

Exam Traps

1-Minute Recap

"Okay, let’s lock this in. Limiting reagent problems are just three steps: 1. Convert everything to moles—mass to moles, or molarity × volume to moles. 2. Pick one reactant and calculate how much of the other you’d need using the mole ratio from the balanced equation. 3. Compare what you need to what you have—if you need more than you have, the other reactant is limiting.

Then, use the limiting reagent to calculate the product. That’s it. No shortcuts—always follow the steps. And watch out for traps: different units, excess reagent questions, and sneaky mole ratios. You’ve got this. Now go ace that exam!