2Al + 3H2SO4 → Al2(SO4)3 + 3H2
\(n_{Al}=\frac{8,1}{27}=0,3\left(mol\right)\)
a) Theo PT: \(n_{H_2SO_4}=\frac{3}{2}n_{Al}=\frac{3}{2}\times0,3=0,45\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,45\times98=44,1\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\frac{44,1}{12,25\%}=360\left(g\right)\)
b) Theo pT: \(n_{H_2}=n_{H_2SO_4}=0,45\left(mol\right)\)
\(\Rightarrow m_{H_2}=0,45\times2=0,9\left(g\right)\)
Ta có: \(m_{dd}saupư=8,1+360-0,9=367,2\left(g\right)\)
Theo Pt: \(n_{Al_2\left(SO_4\right)_3}=\frac{1}{2}n_{Al}=\frac{1}{2}\times0,3=0,15\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,15\times342=51,3\left(g\right)\)
\(\Rightarrow C\%_{Al_2\left(SO_4\right)_3}=\frac{51,3}{367,2}\times100\%=13,97\%\)
\(n_{Al}=\frac{m}{M}=\frac{8,1}{27}=0,3\left(mol\right)\\ PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
(mol) 2 3 1 3
(mol) 0,3 0,45 0,15 0,45
\(m_{H_2SO_4}=n.M=0,45.98=44,1\left(g\right)\\ \rightarrow m_{ddH_2SO_4}=\frac{44,1.100}{12,25}=360\left(g\right)\)
\(C\%_{ddM}=\frac{0,15.342}{8,1+360-0,45.2}.100\%=13,97\left(\%\right)\)
