\(a,n_{Al}=\dfrac{6,75}{27}=0,25(mol)\\ PTHH:2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow n_{AlCl_3}=n_{Al}=0,25(mol)\\ \Rightarrow m_{AlCl_3}=0,25.133,5=33,375(g)\\ b,n_{H_2}=1,5.n_{Al}=0,375(mol)\\ \Rightarrow V_{H_2}=0,375.22,4=8,4(l)\\ \Rightarrow V_{H_2(tt)}=8,4.90\%=7,56(l)\)
\(n_{Al}=\dfrac{m}{M}=\dfrac{6,75}{27}=0,25\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(\Rightarrow n_{AlCl_3}=n_{Al}=0,25\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=n.M=0,25.133,5=33,375\left(g\right)\)
\(b.n_{H_2}=\dfrac{3}{2}.n_{Al}=1,5.0,25=3,75\left(mol\right)\\ \Rightarrow V_{H_2}=n.22,4=3,75.22,4=84\left(l\right)\)
