$a,PTPƯ:2Al+3H_2SO_4\xrightarrow{} Al_2(SO_4)_3+3H_2↑$
$b,n_{Al}=\dfrac{7,1}{27}=\dfrac{71}{270}mol.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{71}{180}mol.$
$⇒V_{H_2}=\dfrac{71}{180}.22,4=8,8356mol.$
$c,Theo$ $pt:$ $n_{H_2SO_4}=\dfrac{3}{2}n_{Al}=\dfrac{71}{180}mol.$
Đổi 500 ml = 0,5 lít.
$⇒CM_{H_2SO_4}=\dfrac{\frac{71}{180}}{0,5}=0,1972l=197,2ml.$
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