Ta có mH2SO4 = V . D = 84,74 . 1,143 = 96,85782 ( gam )
=> nH2SO4 = \(\dfrac{\text{96,85782}\times2}{98}\) \(\approx\) 0,2 ( mol )
2R + 3H2SO4 \(\rightarrow\) R2(SO4)3 + 3H2
\(\dfrac{2}{15}\)......0,2
=> \(\dfrac{2}{15}\) = \(\dfrac{3,6}{M_R}\)
=> MR = 27
=> R là Al