# How would you determine the limiting reactant if given 50.00 ml of 0.250 M Pb(NO3)2 reacting with 25.00 ml of 0.350 M CaCl2?

Jan 7, 2016

$C a C {l}_{2} \left(a q\right) + P b {\left(N {O}_{3}\right)}_{2} \left(a q\right) \rightarrow C a {\left(N {O}_{3}\right)}_{2} \left(a q\right) + P b C {l}_{2} \left(s\right) \downarrow$

#### Explanation:

The stoichiometric equation is given above. But chemistry is an experimental science. You simply have to know that all nitrates are soluble in water, and most halides are soluble in water, EXCEPT for silver chloride, lead chloride, and mercurous chloride ($H {g}_{2} C {l}_{2}$).

Lead chloride is insoluble and will precipitate from solution as a solid.

Apologies for this spray. It had to be said before I embarked on the solution.

So, from the equation, 1 equiv of calcium chloride will react with 1 equiv of lead nitrate. So (finally!) we work out the quantities of the metal reagents.

Moles of lead nitrate: $50.00 \times {10}^{- 3} \cancel{L} \times 0.250 \cdot m o l \cdot \cancel{{L}^{- 1}}$ $=$ ??mol.

Moles of calcium chloride: $25.00 \times {10}^{- 3} \cancel{L} \times 0.350 \cdot m o l \cdot \cancel{{L}^{- 1}}$ $=$ ??mol.

Clearly, the lead ion will be in excess, but you will have to do the calculation.