Ta có:
\(m_{HCl}=\frac{20.124,1}{100}=24,82\left(g\right)\)
\(\Rightarrow n_{HCl}=\frac{24,82}{36,5}=0,68\left(mol\right)\)
Gọi a,b lần lượt là số mol Al,CaCO3 phản ứng
\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\)
_______a ______3a________a _________1,5a
\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
b___________2b ______ b_______b
a)Giải hệ PT:
\(\left\{{}\begin{matrix}3a+2b=0,68\left(1\right)\\3a+44b=7,4\left(2\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,12\\b=0,16\end{matrix}\right.\)
\(\Rightarrow m_{Al}=0,12.27=3,24\left(g\right)\)
\(\%m_{Al}=\frac{3,24.100}{3,24+16}=16,84\%\)
\(\%_{CaCO3}=100\%-16,84\%=83,16\%\)
b)\(m_{Dd\left(spu\right)}=3,24+16+124,1-7,4=135,94\left(g\right)\)
\(C\%_{AlCl3}=\frac{0,12.133,5.100}{135,94}=11,78\%\)
\(C\%_{CaCl2}=\frac{0,16.111.100}{135,94}=13,06\%\)
