a)
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{73.20}{100.36,5}=0,4\left(mol\right)\)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
ban đầu 0,1 0,4
sau pư 0 0,1 0,1 0,3
b)
\(m_{dd}=2,7+73-0,3.2=75,1\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{0,1.133,5}{75,1}.100\%=17,78\%\\C\%_{HCl.dư}=\dfrac{0,1.36,5}{75,1}.100\%=4,86\%\end{matrix}\right.\)
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
\(m_{HCl}=\dfrac{73.20\%}{100\%}=14,6\left(g\right)\)
\(n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,1 --- > 0,3 --- > 0,1 --- > 0,15 (mol)
LTL : \(\dfrac{0,1}{2}< \dfrac{0,4}{6}\)
=> HCl dư sau pứ
\(m_{H_2}=0,15.2=0,3\left(g\right)\)
\(mAlCl_3=0,1.133,5=13,35\left(g\right)\)
\(m_{dd}=mAl+mddHCl-mH_2=2,7+73-0,3=75,4\left(g\right)\)
\(m_{HCl\left(sau.pứ\right)}=\left(0,4-0,3\right).36,5=3,65\left(g\right)\)
=> \(C\%_{AlCl_3}=\dfrac{13,35.100}{75,4}=17,71\%\)
\(C\%_{HCl}=\dfrac{3,65.100}{75,4}=4,84\%\)
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{73.20}{\dfrac{100}{36,5}}=0,4\left(mol\right)\\ pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ LTL:\dfrac{0,1}{2}< \dfrac{0,4}{6}\)
=> HCl dư
\(m_{\text{dd}\left(saup\text{ư}\right)}=2,7+73-0,15.2=75,4\left(g\right)\\ C\%_{AlCl_3}=\dfrac{0,1.133,5}{75,4}.100\%=17,7\%\\ n_{HCl\left(d\text{ư}\right)}=0,4-0,3=0,1\left(mol\right)\\ C\%_{HCl\left(d\text{ư}\right)}=\dfrac{0,1.36,5}{75,4}.100\%=4,84\%\)
