Ta có:
\(m_H:m_S:m_O=0,2:3,2:6,4\\ \rightarrow n_H:n_S:n_O=\dfrac{0,2}{1}:\dfrac{3,2}{32}:\dfrac{6,4}{16}=1:1:4\)
\(CTPT:\left(H_2SO_4\right)_n\)
Mà M = 98 (g/mol)
=> n = 1
CTHH: H2SO4
\(n_{Al}=\dfrac{0,54}{27}=0,02\left(mol\right)\)
PTHH:
2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2
0,02 0,01
\(\rightarrow m_{Al_2\left(SO_4\right)_3}=0,01.342=3,42\left(g\right)\)