Ta có: \(n_{Al}=\dfrac{12,15}{27}=0,45\left(mol\right)\)
\(m_{HCl}=\dfrac{109,5.20}{100}=21,9\left(g\right)\\ =>n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)
PTHH: 2Al + 6HCl -> 2AlCl3 + 3H2
Theo PTHH và đề bài, ta có:
\(\dfrac{n_{Al\left(đề\right)}}{n_{Al\left(PTHH\right)}}=\dfrac{0,45}{2}>\dfrac{n_{HCl\left(đề\right)}}{n_{HCl\left(PTHH\right)}}=\dfrac{0,6}{6}\)
=> Al dư, HCl hết tính theo HCl
=> \(n_{Al\left(phảnứng\right)}=\dfrac{2.0,6}{6}=0,2\left(mol\right)\\ =>n_{Al\left(dư\right)}=0,45-0,2=0,25\left(mol\right)\\ =>m_{Al\left(dư\right)}=0,25.27=6,75\left(g\right)\)
b) Dung dịch sau phản ứng là AlCl3.
Ta có: \(n_{H_2}=\dfrac{3.0,6}{6}=0,3\left(mol\right)\)
=> \(m_{H_2}=0,3.2=0,6\left(g\right)\)
\(n_{AlCl_3}=\dfrac{2.0,6}{6}=0,2\left(mol\right)\\ =>m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
Theo ĐLBTKL, ta có:
\(m_{ddAlCl_3}=m_{Al\left(banđầu\right)}-m_{Al\left(dư\right)}+m_{ddHCl}-m_{H_2}\\ =12,15-6,75+109,5-0,6=114,3\left(g\right)\)
=> \(C\%_{ddsauphảnứng}=\dfrac{m_{AlCl_3}}{m_{ddAlCl_3}}.100\%=\dfrac{26,7}{114,3}.100\approx23,36\%\)
