Ta có:
\(n_{khí}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Mg}:a\left(mol\right)\\n_{MgCO3}:b\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow24a+84b=13,6\)
a, \(PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\left(1\right)\)
\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)
Theo PTHH: (1), (2) :
\(n_{khi}=n_{Mg}+n_{MgCO3}=a+b=0,3\left(mol\right)\)
Giải hệ PT:
\(\left\{{}\begin{matrix}24a+84b=13,6\\a+b=0,3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,193\\b=0,107\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\frac{0,193.24}{13,6}.100\%=34,06\%\\\%m_{MgCO3}=100\%-34,06\%=65,94\%\end{matrix}\right.\)
b,
Theo PTHH (1), (2):
\(n_{HCl}=2n_{Mg}+2n_{MgCO3}=2.0,193+2.0,107=0,6\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)
\(C\%=\frac{m_{ct}}{m_{dd}}.100\%\Rightarrow m_{dd}=\frac{m_{ct}.100}{C\%}1=109,5\)
c, \(m_{dd\left(sau\right)}=m_{hh}+m_{dd\left(HCl\right)}-m_{khí}\)
\(\Rightarrow m_{dd\left(sau\right)}=13,6+109,5-2.0,193-44.0,107=118,006\left(g\right)\)
\(n_{MgCl2}=0,3\left(mol\right)\Rightarrow m_{MgCl2}=28,5\left(g\right)\)
\(C\%_{MgCl2}=\frac{28,5}{118,006}.100\%=24,15\%\)
