Stoichiometry and Limiting Reagent

A balanced equation obeys the law of conservation of mass — atoms are neither created nor destroyed, so the same atoms on the left must reappear on the right in fixed whole-number ratios. For a general reaction

where: $a, b, c, d$ are the balancing coefficients, and $A, B$ are reactants forming products $C, D$. The coefficients are a mole ratio, not a mass ratio, so every calculation must start by converting given masses to moles:

where: $n$ is amount of substance in mol, $m$ is mass in grams, and $M$ is molar mass in $\text{g mol}^{-1}$. Once both reactants are expressed in moles, the balanced equation lets us convert from one species to any other using its coefficients as a ratio. When two reactant quantities are given, they are rarely present in the exact mole ratio the equation demands. Dividing each reactant's moles by its own coefficient and comparing the results tells us which one is used up first:

where: the smaller quotient identifies the limiting reagent — the reactant that runs out first and therefore caps the maximum amount of product. The reactant with the larger quotient is left over as the excess reagent. Every "how much product forms" calculation must then use the limiting reagent's moles only. This idea matters beyond the exam hall. An industrial chemist mixing reagents on a plant scale, or a student titrating an unknown solution, always has to ask which substance runs out first, because that is what fixes the maximum possible yield — no amount of the excess reagent can push the product beyond what the limiting reagent allows. Once the limiting reagent is fixed, finding how much of the excess reagent remains is straightforward: compute how many moles of it the reaction actually consumes, using the mole ratio to the limiting reagent, and subtract that from the moles originally supplied. The leftover amount can then be converted back to a mass with $m = n \times M$ if the question asks for it.

Question. $8\ \text{g}$ of $\text{H}_2$ is reacted with $28\ \text{g}$ of $\text{N}_2$ according to $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$. Identify the limiting reagent and find the mass of $\text{NH}_3$ formed, along with the mass of excess reagent remaining.

Step 1. Convert both masses to moles. $n(\text{N}_2) = 28/28 = 1\ \text{mol}$, using $M(\text{N}_2) = 28\ \text{g mol}^{-1}$. $n(\text{H}_2) = 8/2 = 4\ \text{mol}$, using $M(\text{H}_2) = 2\ \text{g mol}^{-1}$.

Step 2. Divide by the coefficient of each reactant. $n(\text{N}_2)/1 = 1$ and $n(\text{H}_2)/3 = 4/3 = 1.33$. The smaller quotient, $1$, belongs to $\text{N}_2$, so $\text{N}_2$ is the limiting reagent.

Step 3. All $1\ \text{mol}$ of $\text{N}_2$ reacts, consuming $1 \times 3 = 3\ \text{mol}$ of $\text{H}_2$ out of the $4\ \text{mol}$ supplied. Excess $\text{H}_2$ remaining is $4 - 3 = 1\ \text{mol}$, which is $1 \times 2 = 2\ \text{g}$.

Step 4. Moles of $\text{NH}_3$ formed, based on the limiting reagent: $n(\text{NH}_3) = 1\ \text{mol}\ \text{N}_2 \times \dfrac{2}{1} = 2\ \text{mol}$. Mass: $m(\text{NH}_3) = 2 \times 17 = 34\ \text{g}$, using $M(\text{NH}_3) = 14 + 3(1) = 17\ \text{g mol}^{-1}$.

$\text{N}_2$ is the limiting reagent; $34\ \text{g}$ of $\text{NH}_3$ forms, with $2\ \text{g}$ ($1\ \text{mol}$) of $\text{H}_2$ left unreacted.

Competitive stretch

If instead $56\ \text{g}$ of $\text{N}_2$ were supplied with the same $8\ \text{g}$ of $\text{H}_2$, then $n(\text{N}_2) = 2\ \text{mol}$ so $n(\text{N}_2)/1 = 2$, while $n(\text{H}_2)/3$ is still $1.33$. Now $\text{H}_2$ is limiting: all $4\ \text{mol}$ reacts, consuming only $4/3 = 1.33\ \text{mol}$ of the $2\ \text{mol}\ \text{N}_2$ available, and $n(\text{NH}_3) = 4 \times \dfrac{2}{3} = 2.67\ \text{mol}$, giving $m(\text{NH}_3) = 2.67 \times 17 = 45.3\ \text{g}$. Doubling one reactant's mass can flip which reagent is limiting — always recheck both quotients rather than assuming the answer stays the same. The leftover $\text{N}_2$ in this second case is $2 - 1.33 = 0.67\ \text{mol}$, or $0.67 \times 28 = 18.7\ \text{g}$, found the same way as before: subtract the moles actually consumed from the moles originally supplied.

• Using given masses directly in the mole ratio. Coefficients relate moles, not grams — always convert with $n = m/M$ before comparing or reacting quantities.
• Assuming the larger mass is automatically in excess. A reactant can weigh more yet still be limiting if its molar mass is large or its coefficient is small; only $n/\text{coefficient}$ decides.
• Forgetting to balance the equation first. An unbalanced equation gives the wrong mole ratio, and every downstream mass or mole calculation inherits the error.
• Calculating product from the excess reagent. Once the limiting reagent is identified, product amount must be based on its moles alone — using the leftover reagent overestimates the yield.

Ask the tutor: Want to try identifying the limiting reagent in a three-reactant problem, or check your own mass-to-mass calculation step by step?