$PTHH : \\2Na+2H_2O\to 2NaOH+H_2↑ \\2K+2H_2O\to 2KOH+H_2↑ \\Ba+2H_2O\to Ba(OH)_2+2H_2↑ \\n_{H_2}=\dfrac{2,24}{22,4}=0,1mol \\⇒m_{H_2}=0,1.2=0,2g$
$\texttt{Theo pt :}$
$n_{H_2O}=2.n_{H_2}=2.0,1=0,2mol$
$⇒m_{H_2O}=0,2.18=3,6g$
$\texttt{Theo ĐLBTKL :}$
$m=m_{rắn}=9,95+3,6-0,2=13.35g$