a) PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
Ta có: \(n_{Al}=\dfrac{4,05}{27}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{AlCl_3}=0,15\left(mol\right)\\n_{H_2}=0,225\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,225\cdot22,4=5,04\left(l\right)\\m_{AlCl_3}=0,15\cdot133,5=20,025\left(g\right)\end{matrix}\right.\)
b) PTHH: \(PbO+H_2\xrightarrow[]{t^o}Pb+H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{H_2}=0,225\left(mol\right)\\n_{PbO}=\dfrac{66,9}{223}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) PbO còn dư, tính theo H2
\(\Rightarrow\left\{{}\begin{matrix}n_{Pb}=0,225\left(mol\right)\\n_{PbO}=0,075\left(mol\right)\end{matrix}\right.\) \(\Rightarrow m_{rắn}=0,225\cdot207+0,075\cdot223=63,3\left(g\right)\)