PTHH: \(Na_2SO_3+2HCl\rightarrow2NaCl+H_2O+SO_2\) (1)
\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\) (2)
a) Ta có: \(\overline{M}_A=1,75\cdot32=56\)
Theo phương pháp đường chéo: \(\frac{n_{CO_2}}{n_{SO_2}}=\frac{3}{2}\) \(\Rightarrow\left\{{}\begin{matrix}\%V_{CO_2}=60\%\\\%V_{SO_2}=40\%\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}V_{CO_2}=6,72\left(l\right)\\V_{SO_2}=4,48\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{CO_2}=\frac{6,72}{22,4}=0,3\left(mol\right)\\n_{SO_2}=\frac{4,48}{22,4}=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Na_2SO_3}=0,2mol\\n_{CaCO_3}=0,3mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Na_2SO_3}=0,2\cdot126=25,2\left(g\right)\\m_{CaCO_3}=0,3\cdot100=30\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CaCO_3}=\frac{30}{25,2+30}\cdot100\approx54,35\%\\\%m_{Na_2SO_3}=45,65\%\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}n_{HCl\left(1\right)}=0,4mol\\n_{HCl\left(2\right)}=0,6mol\end{matrix}\right.\)
\(\Rightarrow n_{HCl}=1mol\) \(\Rightarrow m_{HCl}=36,5\left(g\right)\)
Mặt khác: \(\left\{{}\begin{matrix}m_{SO_2}=0,2\cdot64=12,8\left(g\right)\\m_{CO_2}=0,3\cdot44=13,2\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{dd}=m_G+m_{ddHCl}-m_A=25,2+30+250-12,8-13,2=279,2\left(g\right)\)\(\Rightarrow C\%_{HCl}=\frac{36,5}{279,2}\cdot100\approx13,07\%\)
