$a,PTPƯ:Fe+2HCl\xrightarrow{} FeCl_2+H_2↑$
$n_{H_2}=\dfrac{2,24}{22,4}=0,1mol.$
$Theo$ $pt:$ $n_{Fe}=n_{H_2}=0,1mol.$
$⇒m_{Fe}=0,1.56=5,6g.$
$Theo$ $pt:$ $n_{HCl}=2n_{H_2}=0,2mol.$
$⇒m_{HCl}=0,2.36,5=7,3g.$
$b,PTPƯ:2H_2+O_2\xrightarrow{t^o} 2H_2O$
$Theo$ $pt:$ $n_{H_2O}=n_{H_2}=0,1mol.$
$⇒m_{H_2O}=0,1.18=1,8g.$
$⇒V_{H_2O}=\dfrac{1,8}{1}=1,8ml.$
chúc bạn học tốt!
