Ta có PTHH
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
\(m_{H_2SO_4}=\frac{14,6\times150}{100}=21,9\left(g\right)\Rightarrow n_{H_2SO_4}=\frac{219}{980}\left(mol\right)\)
\(n_{Al}=\frac{2}{3}n_{H_2SO_4}=\frac{73}{490}\left(mol\right)\Rightarrow m_{Al}=\frac{73}{490}\times27=\frac{1971}{490}\left(g\right)\)
\(m_{HCl}=\frac{9,8\times150}{100}=14,7\left(g\right)\Rightarrow n_{HCl}=\frac{147}{365}\left(mol\right)\)
\(\Rightarrow n_{Al}=\frac{1}{3}n_{HCl}=\frac{49}{365}\left(mol\right)\Rightarrow m_{Al}=\frac{49}{365}\times27=\frac{1323}{365}\left(g\right)\)
\(\Rightarrow m=m_{Al\left(1\right)}+m_{Al\left(2\right)}=\frac{1971}{490}+\frac{1323}{365}\approx7,65\left(g\right)\)
Có
\(n_{Al_2\left(SO_4\right)_3}=\frac{1}{3}n_{H_2SO_4}=\frac{73}{980}\left(mol\right)\Rightarrow C\%\left(Al_2\left(SO_4\right)_3\right)=\frac{\frac{73}{980}\times342}{150+7,65}.100\%\approx4,04\%\)
\(n_{AlCl_3}=\frac{1}{3}n_{HCl}=\frac{49}{635}\left(mol\right)\Rightarrow C\%\left(AlCl_3\right)=\frac{\frac{49}{635}}{150+7,65}.100\%\approx6,53\%\)
hình như em ghi ngược nồng độ 2 axit hay sao ấy
