How many moles of lead (II) hydroxide will be produced from 3.0 moles of lithium hydroxide assuming there is an excess of lead (II) sulfate?

Mar 19, 2015

Zero moles of lead (II) hydroxide will be produced by this reaction because no reaction takes place.

Lead (II) sulfate, or $P b S {O}_{4}$ is Insoluble in aqueous solution, which means that it will not dissociate to allow a reaction to take place.

Lithium hydroxide, or $L i O H$, will dissociate into $L {i}^{+}$ cations and $O {H}^{-}$ anions, but there will be no counterions for them to react with.

$L i O {H}_{\left(a q\right)} + P b S {O}_{4 \left(s\right)} \to \text{No reaction}$

This being said, I will show you how to figure out the number of moles produced assuming that a reaction would take place.

So, every time you're asked to figure out how many moles (or grams, for that matter) of something are produced from something else, your tool of choice is the mole ratio.

$2 L i O H + P b S {O}_{4} \to L {i}_{2} S {O}_{4} + P b {\left(O H\right)}_{2}$
Notice that you have a $\text{2:1}$ mole ratio between lithium hydroxide and lead (II)) hydroxide - this means that, regardless of how many moles of the former react, you'll always produce 2 times fewer moles of the latter.
$\text{3.0 moles LiOH" * ("1 mole"Pb(OH)_2)/"2 moles LiOH" = "1.5 moles}$ $P b {\left(O H\right)}_{2}$