# Question #ccabf

Nov 23, 2015

Each Mol of $N a O H$ will produce half a Mol of $N i {\left(O H\right)}_{2}$

#### Explanation:

Because
$N a O H \to N {a}^{+} + O {H}^{-}$ and
$N {i}^{2 +} + 2 O {H}^{-} \to N i {\left(O H\right)}_{2}$

Total Mols of $N a O H$:
$60 m L \cdot 0.45 M o l / L = 27 m M o l$ of $N a O H$
(dont forget the $m$ for milli=one thousandth)

This will produce $\frac{27}{2} = 13.5 m M o l$ of $N i {\left(O H\right)}_{2}$

Now you multiply this by the molecular mass number to get the weight in milligrams (divide by 1000 to get the grams).

Nickel(II)hydroxide may be formed without or (more likely) with crystal water (monohydrate: $N i {\left(O H\right)}_{2.} {H}_{2} O$)

In the first case you multiply by 92.7, in the other case by 110.7