# How many milliliters of 0.200M NaOH solution are needed to react with 22.0 mL of a 0.490 M NiCl_2 solution?

Dec 8, 2015

There are $0.0108$ $m o l$ $N i C {l}_{2}$. $0.0216$ $m o l$ $N a O H$ are required, given $0.200$ $m o l \cdot {L}^{-} 1$ $N a O H \left(a q\right)$.

#### Explanation:

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

There are $0.0108$ $m o l$ $N i C {l}_{2}$ ($22.0 \times {10}^{- 3} \cancel{L} \times 0.490 \cdot m o l \cdot \cancel{{L}^{-} 1}$). So, by stoichiometry $0.0216$ $m o l$ $N a O H$ are required.

There is $0.200$ $m o l \cdot {L}^{-} 1$ $N a O H$ solution available, so...

$\frac{0.0216 \cdot \cancel{m o l}}{0.200 \cdot \cancel{m o l} \cdot \cancel{{L}^{-} 1}} \times 1000 \cdot m L \cdot \cancel{{L}^{-} 1}$ $=$ ?? mL