cho 300g dd CuCl ² 15% tác dụng với 200g dd NaOH.Tính nồng độ % các chất có trong dd sau phản ứng

$n_{CuCl_2}=\dfrac{300.15\%}{100\%.135}=\dfrac{1}{3}(mol)\\ PTHH:CuCl_2+2NaOH\to Cu(OH)_2\downarrow+2NaCl\\ \Rightarrow n_{Cu(OH)_2}=n_{CuCl_2}=\dfrac{1}{3}(mol);n_{NaCl}=2n_{CuCl_2}=\dfrac{2}{3}(mol)\\ \Rightarrow m_{Cu(OH)_2}=\dfrac{1}{3}.98=32,67(g);m_{NaCl}=\dfrac{2}{3}.58,5=39(g)\\ \Rightarrow C\%_{NaCl}=\dfrac{39}{300+200-32,67}.100\%=8,345\%$

$\text{m$_{CuCl2}$=300.15%=45(g)}$

$\text{⇒n$_{CuCl2}$=45:135=$\frac{1}{3}$ (mol)}$

$\text{PTHH:}$

$\text{CuCl2+2NaOH→Cu(OH)2↓+2NaCl}$

$\text{$\frac{1}{3}$                 →$\frac{1}{3}$  →$\frac{2}{3}$(mol)}$

$\text{⇒m$_{dd sau p/ứ}$=300+200-$\frac{1}{3}$.98≈467,3(g)}$

$\text{⇒%m$_{NaCl}$=$\frac{\frac{2}{3}.58,5}{467,3}$.100% ≈8,35%}$