a) 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
b)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
0,2<-----0,3<------------------0,3
=> mAl = 0,2.27 = 5,4 (g)
c) \(m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)
=> \(m_{dd.H_2SO_4}=\dfrac{29,4.100}{30}=98\left(g\right)\)