# How many moles of  Pb(NO_3)_2 are required if 12 moles of  Al(NO_3)_3 are produced?

$3 P b {\left(N {O}_{3}\right)}_{2} \left(a q\right) + 3 A {l}_{2} {\left(S {O}_{4}\right)}_{3} \left(a q\right) \rightarrow 3 P b S {O}_{4} \left(s\right) \downarrow + 6 A l {\left(N {O}_{3}\right)}_{3} \left(a q\right)$
So if $12 \cdot m o l$ of aluminum nitrate are produced, clearly, from the stoichiometry of the given reaction, $6$ $m o l$ of $\text{lead nitrate}$ were required. But don't trust my arithmetic!