the application of stoichiometric calculations to reactions in aqueous solution, including the use of n = cV and balanced equations to determine limiting reagent, mass or concentration of reactants and products, and percentage yield where appropriate

What this dot point is asking

VCAA wants you to do stoichiometric calculations for reactions in aqueous solution: use $n = cV$ to find moles of each reactant, use the balanced equation to find the mole ratio, identify the limiting reagent, calculate the mass or concentration of products, and where appropriate calculate percentage yield.

The answer

The core formula

For any solute in solution:

where $n$ is moles, $c$ is concentration in $mol\ L^{-1}$, and $V$ is volume in litres. If you have $V$ in millilitres, either convert to litres or use $c$ in $mol\ mL^{-1}$ (rare; convert).

For solids and pure liquids the corresponding formulas are $n = m/M$ (with $m$ in grams and $M$ in $g\ mol^{-1}$) and, for gases at standard conditions, $n = V/V_m$. In Unit 2 most of the work is done with $n = cV$.

The stoichiometry recipe for aqueous reactions

1. Write the balanced equation with state symbols.
2. Calculate the moles of each reactant using $n = cV$ (or $n = m/M$ if a solid is being added to solution).
3. Identify the limiting reagent by dividing the moles available by the coefficient in the balanced equation. The smaller ratio is the limiting reagent. Alternative: assume one reactant is limiting, calculate the moles of the other needed, compare with what is present.
4. Use the mole ratio to find moles of product.
5. Convert to the quantity asked for: mass, concentration, percentage yield, or amount of an excess reagent remaining.

Limiting reagent in detail

When two solutions are mixed, the reactants are rarely present in exact stoichiometric proportions. The limiting reagent runs out first; once it is gone, no more product can form. The other reactant is the excess reagent, and some of it remains in solution after the reaction.

The clearest method: divide moles by the coefficient.

For $aA + bB \to products$:

The smaller of these is the limiting reagent.

Percentage yield

The theoretical yield is the mass (or amount) of product predicted by stoichiometry from the limiting reagent, assuming the reaction goes to completion and nothing is lost.

The actual yield is what is actually measured at the end of the experiment.

Common reasons for yield below 100 percent in solution chemistry:

• The reaction does not go to completion (it is an equilibrium).
• Loss of material during transfer, filtration or drying.
• Side reactions consuming reactant without forming the desired product.
• Solubility of the product allowing some to remain dissolved.
• Incomplete drying or weighing while still wet.

Concentration of a product in solution

When the product stays in solution, the answer is usually a molar concentration:

where $V_{total}$ is the combined volume of the two solutions mixed (assuming volumes are additive, which is a standard VCE approximation for dilute solutions).

A common reaction-type matrix

Worked example

$25.0\ mL$ of $0.200\ mol\ L^{-1}\ HCl$ is added to $35.0\ mL$ of $0.100\ mol\ L^{-1}\ NaOH$. Calculate (a) the limiting reagent, (b) the concentration of the species in excess, and (c) the pH of the final solution. Assume volumes are additive.

(a) $n(HCl) = 0.200 \times 0.0250 = 5.00 \times 10^{-3}\ mol$.
$n(NaOH) = 0.100 \times 0.0350 = 3.50 \times 10^{-3}\ mol$.
Mole ratio is $1:1$. $n(HCl)$ is greater, so $NaOH$ is the limiting reagent.

(b) Moles of $HCl$ left over: $5.00 \times 10^{-3} - 3.50 \times 10^{-3} = 1.50 \times 10^{-3}\ mol$.
Total volume: $25.0 + 35.0 = 60.0\ mL = 0.0600\ L$.
$c(HCl) = 1.50 \times 10^{-3} / 0.0600 = 0.0250\ mol\ L^{-1}$.

(c) $HCl$ is a strong acid, so $[H^+] = 0.0250\ mol\ L^{-1}$.
$pH = -\log(0.0250) = 1.60$.

Common traps

Forgetting to convert mL to L before using $n = cV$. If $c$ is in $mol\ L^{-1}$, $V$ must be in litres.

Assuming the reactant with fewer moles is the limiting reagent. Coefficient matters. If the reaction needs $2A$ per $1B$, then with $n(A) = 0.10$ and $n(B) = 0.10$, $A$ is limiting because you need $0.20\ mol$ of it for that much $B$.

Forgetting to use the combined volume when calculating product concentration after mixing two solutions.

Using the wrong stoichiometric ratio. Always read the coefficients off the balanced equation, not the formulas alone.

Confusing theoretical yield with maximum yield from one reactant. Theoretical yield is from the limiting reagent and the balanced equation.

Forgetting that strong acid and strong base both contribute to $[H^+]$ or $[OH^-]$ after partial neutralisation. Calculate the moles of excess, then divide by total volume.

In one sentence

Stoichiometry of aqueous reactions is the routine of: balanced equation, $n = cV$ for each reactant, find the limiting reagent by coefficient-scaled moles, multiply through the mole ratio for product, then convert to mass, concentration or percentage yield as the question requires.