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Đáp án:

 $\text{C%}$$HCl$$=10$$\text{C%}$

$\text{C%}$$NaCl$$=5,18$$\text{C%}$

Giải thích các bước giải:

Ta có PTHH sau:

$NaOH+HCl$ $\rightarrow$ $NaCl+H_2O$ 

Có $m_{NaOH}$ $=$ $\dfrac{40.10}{100}$ $=4g$

Vậy $n_{NaOH}$ $=$ $\dfrac{4}{40}$ $=0,1$ mol

Có $m_{HCl}$ $=$ $\dfrac{73.15}{100}$ $=10,95g$

Vậy $n_{HCl}$ $=$ $\dfrac{10,95}{36,5}$ $=0,3$ mol

Vì $\dfrac{0,3}{1}$$<$$\dfrac{0,1}{1}$ nên $HCl$ dư

$\rightarrow$ Sau phản ứng thu được $NaCl, HCl$ dư

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Có $n_{HCl}$ dư $=0,3-0,1=0,2$ mol

Vậy $m_{HCl}$ $=$ $0,2.36,5=7,3g$

Vậy $\text{C%}$$HCl$ $=$ $\dfrac{7,3}{73}$.$100=10$$\text{%}$

Dựa vào PTHH) $n_{NaCl}$ $=$ $n_{NaOH}$ $=0,1$ mol

Vậy $m_{NaCl}$ $=0,1.58,5=5,85g$

Có $mdd$sau=$m_{NaOH}$ $+$ $m_{HCl}$ $=40+73=113g$

Vậy $\text{C%}$$NaCl$ $=$ $\dfrac{5,85}{113}$$.100=5,18$$\text{%}$

$\text{m$_{NaOH}$=40.10%=4(g)}$

$\text{⇒n$_{NaOH}$=4:40=0,1(mol)}$

$\text{m$_{HCl}$=73.15%=10,95(g)}$

$\text{⇒n$_{HCl}$=10,95:36,5=0,3(mol)}$

$\text{PTHH:}$

$\text{NaOH+HCl→NaCl+H$_{2}$O}$

$\text{0,1          0,3  (mol)}$

$\text{Lập tỉ lệ :}$

$\text{$\frac{n_{NaOH}}{1}$= $\frac{0,1}{1}$ <$\frac{n_{HCl}}{1}$ =$\frac{0,3}{1}$ }$

$\text{⇒Sau phản ứng HCl dư , NaOH hết}$

$\text{⇒n$_{HCl phản ứng}$=n$_{NaCl}$=n$_{NaOH}$=0,1(mol)}$

$\text{⇒n$_{HCl dư}$=n$_{HCl}$-n$_{HCl phản ứng}$=0,3-0,1=0,2(mol)}$

$\text{⇒m$_{HCl dư }$=0,2.36,5=7,3(g)}$

$\text{m$_{NaCl}$=0,1.58,5=5,85(g)}$

$\text{⇒%m$_{HCl dư}$=$\frac{7,3}{40+73}$.100%≈6,46%}$

$\text{%m$_{NaCl}$=$\frac{5,85}{40+73}$.100%≈5,18%}$