Limiting Reagent & Stoichiometry

4.5 Stoichiometry

Limiting Reagent & Stoichiometry

Slide reactant mole amounts and watch molecules flash and react. The limiting reagent is identified automatically, with theoretical yield and excess calculated live.

Limiting ReagentMole RatiosTheoretical Yield5 Reactions

Topic 4.5

Stoichiometry

Explain changes in the amounts of reactants and products based on the balanced reaction equation for a chemical process.

Because atoms are conserved during a chemical process, you can calculate product amounts from known reactant amounts, or reactant amounts from known product amounts. The coefficients of a balanced equation carry the proportionality information, and those ratios are used with the mole concept.

The universal road map:

given quantity → moles of A → (mole ratio) → moles of B → requested quantity

Everything else is just choosing the right on-ramp and off-ramp. EK 4.5.A.3 makes explicit that stoichiometric calculations combine with the ideal gas law and with molarity, so the entry points are:

From mass: n = m/M
From a solution: n = M × V
From a gas: n = PV/RT

Limiting reactant. When amounts of two or more reactants are given, one runs out first and caps the product. The reliable method: convert each reactant to moles of the same product. The smallest answer is the actual yield-limiting one, and that reactant is the limiting reactant. The other is in excess, and you can find how much is left over by subtracting the amount consumed.

Percent yield. The theoretical yield is what the limiting reactant predicts. The actual yield is what was measured.

percent yield = (actual ÷ theoretical) × 100%

Yields below 100% come from incomplete reaction, competing side reactions, or physical losses during transfer and purification. A yield above 100% means the product was impure or not fully dried.

Key points

The mole ratio from the balanced equation is the only bridge between two different substances.
Find the limiting reactant by converting every reactant to moles of the same product and taking the smallest.
Theoretical yield always comes from the limiting reactant.
Yields above 100% indicate impure or wet product, not a violation of conservation.

Equations

$n = \frac{m}{M}$
on the exam sheet
$n = M \times V$
not on the sheetRearranged from the sheet’s M = n/L.
- $M$ molarity (mol/L)
- $V$ volume of solution (L)
$n = \frac{P V}{R T}$
not on the sheetRearranged ideal gas law.
$% yield = \frac{actual}{theoretical} \times 100%$
not on the sheet

Common mistakes

Comparing masses to find the limiting reactant. Always convert to moles first — grams are not comparable across substances.
Using the wrong mole ratio direction. Write the ratio as a fraction with the unwanted unit on the bottom.
Computing theoretical yield from the excess reactant.
Forgetting to balance the equation first. Every coefficient in the ratio depends on it.

Worked example

25.0 g of N₂ reacts with 6.00 g of H₂ by N₂ + 3 H₂ → 2 NH₃. (a) Identify the limiting reactant. (b) Calculate the theoretical yield of NH₃. (c) If 24.0 g of NH₃ is obtained, find the percent yield. (d) How much excess reactant remains?

(a) Moles and limiting reactant.
n(N₂) = 25.0 ÷ 28.02 = 0.892 mol → ÷1 = 0.892
n(H₂) = 6.00 ÷ 2.016 = 2.976 mol → ÷3 = 0.992

The smaller quotient is N₂, so N₂ is limiting.

(b) Theoretical yield.
0.892 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 1.784 mol NH₃
1.784 × 17.03 g/mol = 30.4 g NH₃

(c) Percent yield.
(24.0 / 30.4) × 100% = 78.9%

(d) Excess H₂ remaining.
H₂ consumed = 0.892 × 3 = 2.676 mol
H₂ left = 2.976 − 2.676 = 0.300 mol → 0.300 × 2.016 = 0.605 g H₂

Full notes for topic 4.5 →