\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(MgCO_3+2HCl\rightarrow MgCl_2+H_2O+CO_2\)
a, \(n_A=0,5\left(mol\right)=n_{H_2}+n_{CO_2}\)
\(m_{hh}=33=24n_{Mg}+84n_{MgCO_3}=84n_{CO_2}+24n_{H_2}\)
\(\Rightarrow\left\{{}\begin{matrix}n_{H_2}=0,15\left(30\%\right)\\n_{CO_2}=0,35\left(70\%\right)\end{matrix}\right.\) ( mol )
b, Có \(n_{HCl}=2n_{H_2}+2n_{CO_2}=1\left(mol\right)\)
Mà \(n_{HClbd}=1,6\left(mol\right)\)
\(\Rightarrow n_{HCldu}=0,6\left(mol\right)\)
Lại có : \(m_{ddsau}=m_{hh}+m_{ddHCl}-m_{hhkhi}=33+880-0,15.2-0,35.44=897,3\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,5.\left(24+71\right)}{897,3}.100\%\approx5,3\%\\C\%_{HCldu}=\dfrac{0,6\left(35,5+1\right)}{897,3}.100\%\approx2,44\%\end{matrix}\right.\)
