2Al + 3H2SO4 \(\rightarrow\) Al2(SO4)3 + 3H2\(\uparrow\)
a) m\(H_2SO_4\) = \(\frac{100.9,8}{100}=9,8\left(g\right)\)
=> n\(H_2SO_4\) = \(\frac{9,8}{98}=0,1\left(mol\right)\)
Theo PT: nAl = \(\frac{2}{3}n_{H_2SO_4}=\frac{2}{3}.0,1=0,067\left(mol\right)\)
=> mAl = 0,067.27 = 1,809(g) = a
Theo PT: n\(H_2\) = n\(H_2SO_4\) = 0,1 (mol)
=> m\(H_2\) = 0,1.2 =0,2 (g)
=> V\(H_2\) = 0,1.22,4 = 2,24 (l) = V
b) Theo PT: n\(Al_2\left(SO_4\right)_3\)= \(\frac{1}{3}n\)\(H_2SO_4\) = \(\frac{1}{3}.0,1=0,03\left(mol\right)\)
=> m\(Al_2\left(SO_4\right)_3\) = 342.0,03 = 10,26 (g)
=> mdd sau pứ = 1,809 + 100 - 0,2 = 101,609 (g)
=> C%\(Al_2\left(SO_4\right)_3\) = \(\frac{10,26}{101,609}.100\%=10,1\%\)
