$a,PTPƯ:Zn+H_2SO_4\xrightarrow{} ZnSO_4+H_2↑$
$b,n_{Zn}=\dfrac{13}{65}=0,2mol.$
$Theo$ $p:$ $n_{H_2SO_4}=n_{H_2}=n_{ZnSO_4}=n_{Zn}=0,2mol.$
$⇒V_{H_2SO_4}=\dfrac{0,2}{1,5}=\dfrac{2}{15}l.$
$c,V_{H_2}=0,2.22,4=4,48l.$
$d,CM_{ZnSO_4}=\dfrac{0,2}{\frac{2}{15}}=1,5M.$
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