Molar concentration connects a measurable solution volume to an amount of solute, so solution reactions can be handled with the same stoichiometric ratios used for any balanced equation.

For a solution of concentration $c$ and volume $V$,

$$n=cV,$$

with $V$ expressed in units consistent with the concentration. The resulting amount can then be converted to another species using the balanced reaction coefficients.

Example

Consider

$$\mathrm{H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O}.$$

What volume of $0.250,\mathrm M$ $\mathrm{NaOH}$ is required to react exactly with $25.0,\mathrm{mL}$ of $0.100,\mathrm M$ $\mathrm{H_2SO_4}$?

First convert the acid volume to liters and find its amount:

$$n_{\mathrm{H_2SO_4}}=(0.100,\mathrm{mol/L})(0.0250,\mathrm L)=2.50\times10^{-3},\mathrm{mol}.$$

The balanced equation requires two moles of sodium hydroxide per mole of sulfuric acid:

$$n_{\mathrm{NaOH}}=(2.50\times10^{-3})\frac{2}{1}=5.00\times10^{-3},\mathrm{mol}.$$

Finally,

$$V_{\mathrm{NaOH}}=\frac{n}{c} =\frac{5.00\times10^{-3}}{0.250} =0.0200,\mathrm L =20.0,\mathrm{mL}.$$

The general path is therefore

$$\text{solution volume}\rightarrow\text{moles of known species}\rightarrow\text{stoichiometric ratio}\rightarrow\text{desired amount or volume}.$$

If several reactants are supplied in finite amounts, the limiting-reactant test must still be applied. Molarity changes how the available amount is calculated; it does not replace ordinary reaction stoichiometry.