Question

Answer 1

a) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

______0,1<----------------------0,05<----0,15______(mol)

=> \(m_{Al}=0,1.27=2,7\left(g\right)\)

b) \(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)