Question

How many grams of Cu(OH)2 will precipitate when excess KOH solution is added to 54.0 mL of 0.572 M CuCl2 solution?

CuCl2 (aq) + 2 KOH (aq) Cu(OH)2 (s) + 2 KCl(aq)

Answer 1

How many grams of $\mathrm{Cu(OH)}_2$ will precipitate when excess KOH solution is added to $54.0~\mathrm{mL}$ of 0.572 M $\mathrm{CuCl}_2$ solution?

$$\mathrm{CuCl}_2(\mathrm{aq}) + 2\mathrm{KOH}(\mathrm{aq}) \rightarrow \mathrm{Cu(OH)}_2(\mathrm{s}) + 2\mathrm{KCl}(\mathrm{aq})$$

$\mathrm{CuCl}_2$ quantity:

$$n_{\mathrm{CuCl}_2} = C_{\mathrm{CuCl}_2} \cdot V_{\mathrm{CuCl}_2} = 0.572 \cdot 0.54 \cdot 10^{-3} = 0.3088 \cdot 10^{-3}~\mathrm{moles}$$

According to reaction stoichiometry $n_{\mathrm{CuCl}_2} = n_{\mathrm{Cu(OH)}_2}$

Then $\mathrm{Cu(OH)}_2$ mass:

$$m_{\mathrm{Cu(OH)}_2} = n_{\mathrm{Cu(OH)}_2} \cdot M_{\mathrm{Cu(OH)}_2} = 0.3088 \cdot 10^{-3} \cdot 97.561 = 0.03~\mathrm{g}$$