Molarity turns a volume of solution into moles, and once you have moles, the mole ratio and the limiting reactant work exactly as before. This is the arithmetic of almost every reaction run in solution.
Key idea
$$n = M \times V \quad (V\ \text{in liters})$$ Change each given volume of solution to moles, use the mole ratio from the balanced (or net ionic) equation, then change to what is asked: a mass, a volume of another solution ($V = n/M$) or a concentration. When two solutions are mixed, check which reactant runs out first.
Why it works
Molarity is moles per liter, so multiplying by liters cancels the "per liter" and leaves moles. Everything after that is the familiar mole road map, with one new road: "liters of solution" connected to "moles" by the molarity. Volumes in milliliters must be divided by 1000 first; otherwise the moles come out 1000 times too big.
Worked example: Volume needed
What volume of $0.100\ \text{M}\ \mathrm{AgNO_3}$ is needed to precipitate all the chloride in $25.0\ \text{mL}$ of $0.150\ \text{M}\ \mathrm{CaCl_2}$? ($\mathrm{Ag^+} + \mathrm{Cl^-} \to \mathrm{AgCl}$)
- Moles of $\mathrm{Cl^-}$: $0.0250\ \text{L} \times 0.150\ \text{M} \times 2 = 7.50 \times 10^{-3}\ \text{mol}$ (two chlorides per $\mathrm{CaCl_2}$).
- Ratio 1 : 1, so $7.50 \times 10^{-3}\ \text{mol}$ of $\mathrm{Ag^+}$.
- $$V = \dfrac{n}{M} = \dfrac{7.50 \times 10^{-3}}{0.100} = 0.0750\ \text{L} = 75.0\ \text{mL}$$
Answer: $75.0\ \text{mL}$
Worked example: Two solutions, one limiting
$50.0\ \text{mL}$ of $0.200\ \text{M}\ \mathrm{BaCl_2}$ is mixed with $40.0\ \text{mL}$ of $0.300\ \text{M}\ \mathrm{Na_2SO_4}$. What mass of $\mathrm{BaSO_4}$ ($233.40\ \text{g/mol}$) forms?
- $\mathrm{Ba^{2+}}$: $0.0500 \times 0.200 = 0.0100\ \text{mol}$. $\mathrm{SO_4^{2-}}$: $0.0400 \times 0.300 = 0.0120\ \text{mol}$.
- They react 1 : 1, so barium (0.0100 mol) is limiting; 0.0020 mol of sulfate is left over.
- Mass: $0.0100 \times 233.40 = 2.33\ \text{g}$.
Answer: $2.33\ \text{g}$
Worked example: Mass of a precipitate
What mass of $\mathrm{Mg(OH)_2}$ ($58.33\ \text{g/mol}$) forms when $30.0\ \text{mL}$ of $0.250\ \text{M}$ NaOH is added to excess $\mathrm{MgCl_2}$ solution? ($\mathrm{Mg^{2+}} + 2\,\mathrm{OH^-} \to \mathrm{Mg(OH)_2(s)}$)
- $n_{\mathrm{OH^-}} = 0.0300 \times 0.250 = 7.50 \times 10^{-3}\ \text{mol}$.
- Two hydroxides per formula unit: $7.50 \times 10^{-3}/2 = 3.75 \times 10^{-3}\ \text{mol}\ \mathrm{Mg(OH)_2}$.
- Mass: $3.75 \times 10^{-3} \times 58.33 = 0.219\ \text{g}$.
Answer: $0.219\ \text{g}$
To find what is left over, subtract the moles used from the moles of the excess reactant, then divide by the total volume of the mixture: in the example, $0.0020\ \text{mol}$ of sulfate in $90.0\ \text{mL}$ is $0.022\ \text{M}$.
Common mistake
Multiplying molarity by milliliters as if they were liters, or skipping the mole ratio. $0.150\ \text{M} \times 25.0 = 3.75$ is not moles; $0.150 \times 0.0250 = 3.75 \times 10^{-3}$ is. And for $\mathrm{CaCl_2}$ the chloride moles are twice the salt's. Check: mL × M gives millimoles, a useful unit if you keep the label.