Bài 1 :
$PTK_{NH_4NO_3} = 80(đvC)$
$\%N = \dfrac{14.2}{80}.100\% = 35\%$
$\%H = \dfrac{1.4}{80}.100\% = 5\%$
$\%O = \dfrac{16.3}{80}.100\% = 60\%$
Bài 2 :
a) $n_{Fe} = \dfrac{2,8}{56} = 0,05(mol)$
$Fe + 2HCl \to FeCl_2 + H_2$
$n_{HCl} =2 n_{Fe} = 0,1(mol)$
$V_{dd\ HCl} = \dfrac{0,1}{2} = 0,05(lít)$
b) $n_{H_2} = n_{Fe} = 0,05(mol)$
$V_{H_2} = 0,05.22,4 = 1,12(lít)$