nBa = \(\dfrac{13,7}{137}\) = 0,1mol
a)Ba + 2H2O -> Ba(OH)2 + H2
0,1 ->0,1 ->0,1
b) VH2 = 0,1. 22,4 = 2,24 (l)
c) mBa(OH)2 = 0,1 . 171 = 17,1 g
=> C% = \(\dfrac{17,1}{13,7+200-0,1.2}\).100% = 8%
d) V = \(\dfrac{17,1}{1,0675}\) = 16 ml = 0,016 (l)
=>CM = \(\dfrac{0,1}{0,016}\) = 6,25 M