OpenSpecEngineering

Finds which of two reactants runs out first, how much product the reaction can make, and how much of the other reactant is left over. Enter the amounts as moles, or as masses with each molar mass worked out from the formula. Use it to plan a batch, size a neutralization or check a yield.

Equations

The page writes the reaction as ν_A A + ν_B B → ν_P P and uses the stoichiometric-table method [1]:

n = m/M     ξ = min(n_A/ν_A, n_B/ν_B)

n_P = ν_P·ξ     n_left = n₀ − ν·ξ

n is an amount in moles, m a mass in grams and M a molar mass in g/mol. ν_A, ν_B and ν_P are the coefficients of the balanced equation. ξ is the extent of reaction: how many times the reaction “happens” as written before one reactant is gone. The reactant with the smaller n/ν is the limiting reagent, and its remainder is zero. With masses, picking a substance fills its molar mass from the formula and the IUPAC atomic weights [2]. Choose Custom to type your own.

Limits and common mistakes

• Balance the equation first. The page trusts the coefficients you enter.
• Compare n/ν, not n. A reactant with more moles can still be limiting if its coefficient is larger.
• The product is the theoretical yield: complete reaction, no side reactions, no losses. Real yields are lower.
• Only one product is tracked. Run the page once per product.
• Use the mass of the pure substance. For a solution such as 37 % HCl, multiply by the mass fraction first.

Worked example

A plant neutralizes 700 g of hydrogen chloride with 1 kg of soda ash: Na₂CO₃ + 2 HCl → 2 NaCl + H₂O + CO₂. Is the soda ash enough, and how much salt forms?

1. Choose Amounts Given As Masses (molar masses from the formula).
2. Choose Reactant A — substance Sodium carbonate (Na₂CO₃). The molar mass fills in as 105.988 g/mol. Enter Mass of Reactant A 1000 g and Stoichiometric Coefficient of A 1.
3. Choose Reactant B — substance Hydrogen chloride (HCl). Its molar mass is 36.458 g/mol. Enter Mass of Reactant B 700 g and Stoichiometric Coefficient of B 2.
4. Choose Product — substance Sodium chloride (NaCl), at 58.44 g/mol, and enter Stoichiometric Coefficient of Product 2.
5. Press Solve. The Solution shows: Limiting Reagent: Reactant A
   Moles of Product Formed: 18.8701 mol
   Reactant B Remaining: 0.330115 mol
   Mass of Product Formed: 1,102.77 g
   Mass of Reactant B Remaining: 12.0353 g

Check by hand: 1000/105.988 = 9.435 mol of soda ash and 700/36.458 = 19.20 mol of HCl. The soda ash can take 2 × 9.435 = 18.87 mol of HCl, which is less than 19.20, so the soda ash is limiting and about 12 g of HCl is left.