Sửa đề cho dễ làm: "200g dd HCl 3,65%"
Ta có: \(n_{Na_2CO_3.10H_2O}=\dfrac{14,3}{106+10\cdot18}=0,05\left(mol\right)\) \(\Rightarrow n_{Na_2CO_3}=0,05\left(mol\right)\)
PTHH: \(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Na_2CO_3}=0,05\left(mol\right)\\n_{HCl}=\dfrac{200\cdot3,65\%}{36,5}=0,2\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,05}{1}< \dfrac{0,2}{2}\) \(\Rightarrow\) Na2CO3 p/ứ hết, HCl còn dư
\(\Rightarrow n_{NaCl}=0,1\left(mol\right)=n_{HCl\left(dư\right)}\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,1\cdot58,5=5,85\left(g\right)\\m_{HCl\left(dư\right)}=0,1\cdot36,5=3,65\left(g\right)\end{matrix}\right.\)
Mặt khác: \(n_{CO_2}=0,05\left(mol\right)\) \(\Rightarrow m_{CO_2}=0,05\cdot44=2,2\left(g\right)\)
\(\Rightarrow m_{dd}=m_{Na_2CO_3.10H_2O}+m_{ddHCl}-m_{CO_2}=212,1\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{5,85}{212,1}\cdot100\%\approx2,76\%\\C\%_{HCl\left(dư\right)}=\dfrac{3,65}{212,1}\cdot100\%\approx1,72\%\end{matrix}\right.\)
