Theo đề ra ta có : nH2 = 1,792/22,4=0,08(mol)
PTHH:
X + H2+O -> X(OH)2 + H2
0,08mol...........0,08mol...0,08mol
a) Ta có : \(M_X=\dfrac{3,2}{0,08}=40=>X:Ca\)
b)nX(OH)2(pư) = 1/2.0,08=0,04(mol)
PTHH :
X(OH)2 + 2HCl -> XCl2 + 2H2O
0,04mol....0,08mol
-> VddHCl = 0,08/2=0,04(l)