Câu 12:

$M_{NO}=14+16=30(g/mol)\\ \%m_N=\dfrac{14}{30}.100\%=46,67\%\\ M_{NH_3}=14+3=17(g/mol)\\ \%m_N=\dfrac{14}{17}.100\%=82,35\%\\ M_{NH_4NO_3}=14.2+4+16.3=80(g/mol)\\ \%m_N=\dfrac{14.2}{80}.100\%=35\%\\ M_{Al(NO_3)_3}=27+(14+16.3).3=213(g/mol)\\ \%m_N=\dfrac{14.3}{213}.100\%=19,72\%\\ M_{Ca(NO_3)_2}=40+(14+16.3).2=164(g/mol)\\ \%m_N=\dfrac{14.2}{164}.100\%=17,07\%$

Câu 13:

$a,M_X=29.1,448\approx 42(g/mol)\\ \text{Đặt CTPT của X là }C_xH_y\\ \%_H=100\%-85,7\%=14,3\%\\ \Rightarrow x:y=\dfrac{85,7\%}{12}:\dfrac{14,3\%}{1}\approx 1:2\\ \Rightarrow CTPT_X:(CH_2)_n\\ \Rightarrow (12+2)n=42\\ \Rightarrow n=3\\ \Rightarrow CTPT_X:C_3H_6\\ b,n_S=\dfrac{4,8}{32}=0,15(mol);n_O=\dfrac{4,8}{16}=0,3(mol)\\ \Rightarrow \dfrac{n_S}{n_A}=1;\dfrac{n_O}{n_A}=2$

Do đó trong 1 mol chất A sẽ có 2 mol O và 1 mol S

Vậy CTHH của A là $SO_2$