$a,PTPƯ:Mg+H_2SO_4\xrightarrow{} MgSO_4+H_2↑$
$n_{Mg}=\dfrac{3,6}{24}=0,15mol.$
$Theo$ $pt:$ $n_{MgSO_4}=n_{Mg}=0,15mol.$
$⇒m_{MgSO_4}=0,15.120=18g.$
$b,Theo$ $pt:$ $n_{H_2}=n_{Mg}=0,15mol.$
$⇒V_{H_2}=0,15.22,4=3,36l.$
$c,Theo$ $pt:$ $n_{H_2SO_4}=n_{Mg}=0,15mol.$
Đổi 140 ml = 0,14 lít.
$⇒CM_{H_2SO_4}=\dfrac{0,15}{0,14}=1,07M.$
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