# Question ecab8

Nov 20, 2017

$4.25$ grams of $C u {\left(O H\right)}_{2}$.

#### Explanation:

The first step is to write the full equation:

$C u C {l}_{2} + 2 N a O H \to C u {\left(O H\right)}_{2} + 2 N a C l$

Using molar quantities of each compound, we can establish that 1 Mole of copper(II) chloride will give 1 Mole of copper(II) hydroxide, when fully reacted.

The amount (in grams) of $C u C {l}_{2}$ we are using can be calculated by using the concentration and the volume.

$\left(\text{56 mL") / ("1000 mL") = 0.056 * ("0.779 M" * M_r " of } C u C {l}_{2}\right)$

The ${M}_{r}$ of $C u C {l}_{2}$ is ${\text{134.5 g mol}}^{-} 1$, so the total amount of available copper(II) chloride is therefore $\text{5.867 g}$.

Using the 1:1 molar ratio of $C u C {l}_{2} : C u {\left(O H\right)}_{2}$, we can work out the final amount of $C u {\left(O H\right)}_{2}$ as the number of moles of copper(II) chloride $\times$ molecular mass of copper(II) hydroxide.

("5.867 g")/("134.5 gmol"^-1) * M_r " of " Cu(OH)_2 = "4.25 g"# of $C u {\left(O H\right)}_{2}$