\(\text{1)}m_{KOH}=40.35\%=14\left(g\right)\\ \rightarrow n_{KOH}=\dfrac{14}{56}=0,25\left(mol\right)\\ PTHH:KOH+HCl\rightarrow KCl+H_2O\\ \text{Theo pthh}:n_{HCl}=n_{KOH}=0,25\left(mol\right)\\ \rightarrow V_{ddHCl}=0,25.0,5=0,125\left(l\right)\)
\(\text{2)}n_{Al}=\dfrac{4,05}{27}=0,15\left(mol\right)\\ n_{H_2SO_4}=200.14,7\%=29,4\left(g\right)\\ \rightarrow n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\\ \text{PTHH}:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\\ \text{LTL}:\dfrac{0,15}{2}< \dfrac{0,3}{3}\rightarrow H_2SO_4\text{ dư}\)
\(\text{Theo pthh}:\left\{{}\begin{matrix}n_{H_2SO_4\left(pư\right)}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,15=0,225\left(mol\right)\\n_{H_2}=n_{H_2SO_4\left(pư\right)}=0,225\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}.0,15=0,075\left(mol\right)\end{matrix}\right.\\ \rightarrow m_{dd\left(\text{sau phản ứng}\right)}=200+4,05-0,3.2=203,45\left(g\right)\)
\(\rightarrow\left\{{}\begin{matrix}C\%_{H_2SO_4\text{ dư}}=\dfrac{\left(0,3-0,225\right).98}{203,45}=3,61\%\\C\%_{Al_2\left(SO_4\right)_3}=\dfrac{342.0,075}{203,45}=12,61\%\end{matrix}\right.\)
