PTHH: 2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2
Ag không phản ứng với HCl vì trong dãy hoạt động hóa học thì Ag đứng sau H.
=> nH2=13,4422,4=0,6nH2=13,4422,4=0,6 (mol)
Theo PTHH: nAl=23nH2=23.0,6=0,4nAl=23nH2=23.0,6=0,4 (mol)
=> mAl = 0,4.27 = 10,8 (g)
=> mAg = 12 - 10,8 = 1,2(g)
2Al + 3H2SO4 → Al2(SO4)3 + 3H2↑
\(n_{H_2}=\frac{13,44}{22,4}=0,6\left(mol\right)\)
a) Theo pT: \(n_{Al}=\frac{2}{3}n_{H_2}=\frac{2}{3}\times0,6=0,4\left(mol\right)\)
\(\Rightarrow m_{Al}=0,4\times27=10,8\left(g\right)\)
\(\Rightarrow m_{Ag}=12-10,8=1,2\left(g\right)\)
\(\%m_{Al}=\frac{10,8}{12}\times100\%=90\%\)
\(\%m_{Ag}=100\%-90\%=10\%\)
b) Theo PT: \(n_{H_2SO_4}=n_{H_2}=0,6\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,6\times98=58,8\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\frac{58,8}{7,35\%}=800\left(g\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\frac{800}{1,025}=780,49\left(ml\right)\)
c) Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\frac{1}{3}n_{H_2}=\frac{1}{3}\times0,6=0,2\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,2\times342=68,4\left(g\right)\)
\(m_{H_2}=0,6\times2=1,2\left(g\right)\)
Ta có: \(m_{dd}saupư=12+800-1,2-1,2=809,6\left(g\right)\)
\(\Rightarrow C\%_{Al_2\left(SO_4\right)_3}=\frac{68,4}{809,6}\times100\%=8,45\%\)
