1/
$a,PTPƯ:2Al+3H_2SO_4\xrightarrow{} Al_2(SO_4)_3+3H_2↑$
$b,n_{Al}=\dfrac{5,4}{27}=0,2mol.$
$Theo$ $pt:$ $n_{Al_2(SO_4)_3}=\dfrac{1}{2}n_{Al}=0,1mol.$
Đổi 350 ml = 0,35 lít.
$⇒CM_{Al_2(SO_4)_3}=\dfrac{0,1}{0,35}=\dfrac{2}{7}M.$
2/
$a,PTPƯ:Fe+H_2SO_4\xrightarrow{} FeSO_4+H_2↑$
$n_{Fe}=\dfrac{11,2}{56}=0,2mol.$
$Theo$ $pt:$ $n_{FeSO_4}=n_{H_2}=n_{H_2SO_4}=n_{Fe}=0,2mol.$
Đổi 150 ml = 0,15 lít.
$⇒CM_{FeSO_4}=\dfrac{0,2}{0,15}=\dfrac{4}{3}M.$
$b,V_{H_2}=0,2.22,4=4,48l.$
$c,CM_{H_2SO_4}=\dfrac{0,2}{0,15}=\dfrac{4}{3}M.$
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