Gọi x là số mol $H_2SO_4$, y là số mol nước trong dd $H_2SO_4$.
Giả sử $m_{dd}=100g$
$\Rightarrow 98x+18y=100$ $(1)$
$m_{H_2}=4,7g$
$\Rightarrow n_{H_2}=\dfrac{4,7}{2}=2,35(mol)$
$2Na+H_2SO_4\to Na_2SO_4+H_2$
$Fe+H_2SO_4\to FeSO_4+H_2$
$Mg+H_2SO_4\to MgSO_4+H_2$
$2Na+2H_2O\to 2NaOH+H_2$
$\Rightarrow x+0,5y=2,35$ $(2)$
$(1)(2)\Rightarrow x=0,248; y=4,2$
$\Rightarrow C\%_{H_2SO_4}=\dfrac{98.0,248.100}{100}=24,304\%$