дано
m(Ba) = 16,44 g
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m(Ba(OH)2)-?
V(H2)-?
2Ba+2HOH-->Ba(OH)2+H2
M(Ba) = 137 g/mol
n(Ba) = m/M = 16.44 / 137 = 0.12 mol
2n(Ba) = n(Ba(OH)2)
n(Ba(OH)2) = 0.12 / 2 = 0.06 mol
M(Ba(OH)2) = 171 g/mol
m(Ba(OH)2) = n*M = 0.06 * 171 = 10.26 g
2n(Ba) = n(H2)
n(H2) = 0.06 mol
V(H2) = n*Vm = 0.06 * 22.4 = 1.344 L
ответ 10.26 г , 1.344 л