$n_{Ca(OH)_2}=\dfrac{9,25}{74}=0,125mol \\n_{SO_2}=\dfrac{2,24}{22,4}=0,1mol \\PTHH :$

                  $Ca(OH)_2 + SO_2\to H_2O+CaSO_3$

$\text{a.Theo pt :   1 mol          1 mol}$

$\text{Theo đbài :  0,125 mol   0,1 mol}$

Ta có tỷ lệ : $\dfrac{0,125}{1}>\dfrac{0,1}{1}$

$\text{⇒Sau pư Ca(OH)2 dư}$

$Theo\ pt : \\n_{Ca(OH)_2\ pư}=n_{SO_2}=0,1mol \\⇒n_{Ca(OH)_2\ dư}=0,125-0,1=0,025mol \\⇒m_{Ca(OH)_2\ dư}=0,025.74=1,85g \\n_{CaSO_3}=n_{H_2O}=n_{SO_2}=0,1mol \\⇒m_{CaSO_3}=0,1.120=12g \\m_{H_2O}=0,1.18=1,8g \\b.PTHH : \\CaO+H_2O\to Ca(OH)_2 \\Theo\ pt : \\n_{CaO}=n_{Ca(OH)_2\ pư}=0,1mol \\⇒m_{CaO}=0,1.56=5,6g \\c.PTHH : \\2Fe+6H_2SO_4\overset{t^o}\to Fe_2(SO_4)_3+6H_2O+3SO_2↑ \\Theo\ pt : \\n_{H_2SO_4}=2.n_{SO_2}=2.0,1=0,2mol \\⇒V_{H_2SO_4}=\dfrac{0,2}{0,4}=0,5l$

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