$a, Cu +2H_2SO_4 \buildrel{t°}\over\to CuSO_4 + SO_2 + 2H_2O$
$n_{\text{Cu}} = \dfrac{9,6}{64}=0,15(\text{mol})$
Theo ptpư : $\to m_{H_2SO_4} = 0,3.98=89,4(\text{gam})$
$b, n_{SO_2} = 0,15 (\text{mol})$
$\Rightarrow V_{SO_2} = 0,15.22,4=3,36 (\text{l})$
$c, 2Fe +6 H_2SO_4 \to Fe_2(SO_4)_3 + 3SO_2 +6 H_2O$
$n_{SO_2} = \dfrac{3,36}{22,4}= 0,15 (\text{mol})$
$\to m_{Fe} = 56.0,1=5,6 (\text{gam})$