\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ \Rightarrow m_{H_2}=0,3.2=0,6\left(g\right)\)
`n_(H_2) = (6,72)/(22,4) = 0,3 (mol)`
`-> m_(H_2) = 0,3 . 2 = 0,6 (g)`
$\rm n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{6,72}{22,4}=0,3(mol)$
`->` $\rm m_{H_2}=n.M=0,3.2=0,6(g)$