Theo đề, ta có: \(n_{HCl}=2.0,3=0,6\left(mol\right)\)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2 (mol) \(\rightarrow\) 0,6 (mol) \(\rightarrow\) 0,2 (mol) \(\rightarrow\)0,3 (mol)
Theo phương trình, ta được:
* \(n_{Al}=\dfrac{1}{3}n_{HCl}=\dfrac{1}{3}.0,6=0,2\left(mol\right)\)
\(\Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\)
* \(n_{H_2}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}.0,6=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
* \(n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
(Vì muối khan thu được chỉ có AlCl3)
Vậy: - \(m=m_{Al}=5,4\left(g\right)\)
- \(V=V_{H_2}=6,72\left(l\right)\)
- \(m_1=m_{AlCl_3}=26,7\left(g\right)\)
