\(n_{H_2}=\dfrac{2,576}{22,4}=0,115\left(mol\right)\)
PTHH: X + aH2O --> X(OH)a + \(\dfrac{a}{2}\)H2
____\(\dfrac{0,23}{a}\)<-----------------------0,115
=> \(M_X=\dfrac{4,6}{\dfrac{0,23}{a}}=20a\left(g/mol\right)\)
Xét a = 1 => MX = 20 (L)
Xét a = 2 => MX = 40(g/mol) => Ca