Revise Reacting masses and volumes (of solutions and gases) for Cambridge 9701 (2.4)

The mole is the SI unit for the amount of a substance. It's the bridge that connects all our calculations. Whether you are given a mass, a gas volume, or a solution concentration, the first step is almost always to convert this quantity into moles. Once you have moles, you can use the stoichiometry of the balanced equation to find the moles of any other substance in the reaction, and then convert back into the desired quantity (mass, volume, etc.).

These calculations often involve reacting a solid to produce a gas, or vice versa. The key is to use the appropriate formula for each substance. For solids, we use mass and molar mass. For gases, we use volume and the molar gas volume.

Moles from mass: $n = \frac{\text{mass}}{M_r}$ \ Moles from gas volume (at RTP): $n = \frac{\text{Volume in dm}^3}{24}$

Titration is a common laboratory technique used to determine the concentration of an unknown solution (the analyte) by reacting it with a solution of known concentration (the titrant). The calculation hinges on finding the moles of the known solution, using the mole ratio to find the moles of the unknown solution, and then calculating its concentration.

Moles from solution: $n = c \times V \quad (\text{where V is in dm}^3)$ \ Concentration from moles: $c = \frac{n}{V} \quad (\text{where V is in dm}^3)$

In most real reactions, reactants are not mixed in perfect stoichiometric amounts. One reactant, the limiting reactant, will be completely used up, causing the reaction to stop. The other reactants are said to be in excess. The amount of product formed is always determined by the initial amount of the limiting reactant. To identify the limiting reactant, calculate the moles of each reactant and compare this to the mole ratio required by the balanced equation.