\(Ba+2H_2O\xrightarrow[]{}Ba\left(OH\right)_2+H_2\\ n_{Ba\left(OH\right)_2}=n_{H_2}=n_{Ba}=0,02mol\\ m_{Ba}=0,02.137=2,74\left(g\right)\\ m_{H_2}=0,02.2=0,04\left(g\right)\\ m_{ddBa\left(OH\right)_2}=100+2,74 -0,04=102,7\left(g\right)\\ m_{Ba\left(OH\right)_2}=0,02.171=3,42\left(g\right)\\ C_{\%Ba\left(OH\right)_2}=\dfrac{3,42}{102,7}\cdot100\approx3,33\%\\ V_{H_2}=0,02.22,4=0,448\left(l\right)\)