All calculatorsTheoretical Yield Calculator
Compute theoretical yield from reactant amounts and stoichiometric coefficients. Identify the limiting reagent, get product moles and (optionally) product grams if you provide its molar mass. See steps and a mini extent chart.
Background
For a balanced reaction, each reactant limits the reaction by the amount of reaction extent it can supply: ξi = ni / νi. The smallest extent is the limiting reagent. Theoretical product moles are nP,th = ξlim · νP. If the product’s molar mass is given, convert to grams.
How this calculator works
- For each reactant, compute moles n (convert grams via n = m/M as needed), then its extent ξ = n / ν.
- The smallest extent is the limiting reagent extent ξlim.
- Theoretical product moles: nP,th = ξlim · νP. If product molar mass is provided, theoretical grams = nP,th · Mₚ.
Formula & Equation Used
Extent per reactant: ξi = ni / νi
Limiting extent: ξlim = min(ξi)
Product moles: nP,th = ξlim · νP
Mass–mole: n = m / M, m = n·M
Example Problems & Step-by-Step Solutions
Example 1 — 2H₂ + O₂ → 2H₂O
Given m(H₂)=4.00 g (M=2.016) → n=1.984 mol; m(O₂)=32.0 g (M=31.998) → n=1.000 mol.
ξ(H₂)=1.984/2=0.992; ξ(O₂)=1.000/1=1.000 ⇒ limiting is H₂ (ξ=0.992).
n(H₂O) = ξ·νₚ = 0.992·2 = 1.984 mol; m(H₂O) with M=18.015 → 35.7 g.
Example 2 — N₂ + 3H₂ → 2NH₃ (all in moles)
n(N₂)=0.50, n(H₂)=1.50 ⇒ ξ(N₂)=0.50/1=0.50; ξ(H₂)=1.50/3=0.50 ⇒ tie.
n(NH₃)=0.50·2=1.00 mol; if M=17.031 → m≈17.03 g.
Example 3 — Fe + S → FeS (1:1; mixed units)
m(Fe)=5.60 g (M=55.845) → n=0.1002 mol; n(S)=0.120 mol.
ξ(Fe)=0.1002/1=0.1002 (limiting), ξ(S)=0.120/1=0.120.
n(FeS)=0.1002·1=0.1002 mol; M≈87.91 → m≈8.80 g.
Frequently Asked Questions
Q: Do I need the product molar mass?
Only if you want theoretical grams; for theoretical moles it’s not required.
Q: What if two reactants tie as limiting?
Then both limit to the same extent; the computed product is the same either way.
Q: Can I enter all reactants in grams?
Yes—just include each molar mass so we can convert to moles.
