0.85 moles formula units of lead nitrate will produce 0.57 moles formula units of chromium (III) nitrate.
Typically, the oxidation state of Pb in lead nitrate tend to be +2. In other words, Pb in lead nitrate tends to exist as [tex]\text{Pb}^{2+}[/tex] ions. The formula for a nitrate ion is [tex]{\text{NO}_3}^{-}[/tex]. The charge on each of the nitrate ion is -1. The charge on the two ions should balance. As a result, each [tex]\text{Pb}^{2+}[/tex] ion in lead nitrate would pair up with two [tex]{\text{NO}_3}^{-}[/tex] ions. The formula for lead nitrate will be [tex]\text{Pb}({\text{NO}_3})_2[/tex]. Each formula unit of lead nitrate will contain one [tex]\text{Pb}^{2+}[/tex] ion and two [tex]{\text{NO}_3}^{-}[/tex] ions.
The "III" in the name "chromium (III) nitrate" is a Roman Numeral. It indicates that the oxidation state of Cr in chromium (III) nitrate is +3. The Cr in that compound will exist as [tex]\text{Cr}^{3+}[/tex]. Similarly, each [tex]\text{Cr}^{3+}[/tex] will pair up with three [tex]{\text{NO}_3}^{-}[/tex] ions. The formula for chromium (III) nitrate will be [tex]\text{Cr}(\text{NO}_3})_3[/tex]. Each formula unit of chromium (III) nitrate will contain one [tex]{\text{NO}_3}^{-}[/tex] ion and three [tex]{\text{NO}_3}^{-}[/tex] ions.
0.85 moles formula units of lead nitrate will contain 0.85 × 2 = 1.7 moles of [tex]{\text{NO}_3}^{-}[/tex] ions. Those nitrate ions will end up in 1.7 / 3 = 0.57 moles formula units of chromium (III) nitrate. As a result, the reaction will produce 0.57 moles formula units of chromium (III) nitrate.
