Bài này khá dài nè, em đăng tách ra nha ^^ 

\(a.\%m_N=\dfrac{14.2}{\left(14+4.1\right).2+32+4.16}.100\approx21,212\%\\ \%m_H=\dfrac{4.2}{\left(14+4.1\right).2+32+4.16}.100\approx6,061\%\\ \%m_S=\dfrac{32}{\left(14+4.1\right).2+32+4.16}.100\approx24,242\%\\ \%m_O=\dfrac{4.16}{\left(14+4.1\right).2+32+4.16}.100\approx48,485\%\)

\(b.m_{N\left(20kg\right)}=20.\dfrac{2.14}{\left(14+4.1\right).2+32+4.16}.100\%\approx4,2424\left(kg\right)\)

Đặt n NH4NO3 = n NH4CL = n (NH4)2SO4 = n (NH4)3PO4 = 1 ( mol )

- Trong 1 mol NH4NO3 có 2 mol N

   => m N = 28 ( g ) =>%m N = 35%

- Trong 1 mol NH4CL có 1 mol N

   => m N = 14 ( g ) => %m N = 26,17%

- Trong 1 mol (NH4)2SO4 có 2 mol N

   => m N = 28 ( g ) => %m N = 21,21%

- Trong 1 mol (NH4)3PO4 có 3 mol N

    => m N = 42 ( g ) => %m N = 28,19%

   Vậy NH4NO3 có hàm lượng N cao nhất

Ta có: MAl2O3 = 27.2+16.3 = 102 g/mol

%Al = .100% = 52,94%

%O = .100% = 47,06%

\(PTK=12.12+1.22+16.11=342< mol>.\)

\(\%C=\dfrac{12.12.100}{342}\approx42,11\%\)

\(\%H=\dfrac{1.22.100}{142}=15,49\%\)

\(\%O=100\%-42,11\%-15,49\%=42,4\%\)

Bài 1:

\(1,M_{MgCO_3}=84(g/mol)\\ \begin{cases} \%_{Mg}=\dfrac{24}{84}.100\%=28,57\%\\ \%_{C}=\dfrac{12}{84}.100\%=14,29\%\\ \%_{O}=100\%-28,57\%-14,29\%=57,14\% \end{cases}\)

\(2,M_{Al(OH)_3}=78(g/mol)\\ \begin{cases} \%_{Al}=\dfrac{27}{78}.100\%=31,62\%\\ \%_{H}=\dfrac{3}{78}.100\%=3,85\%\\ \%_{O}=100\%-31,62\%-3,85\%=64,53\% \end{cases}\)

\(3,M_{(NH_4)_2HPO_4}=132(g/mol)\\ \begin{cases} \%_{N}=\dfrac{28}{132}.100\%=21,21\%\\ \%_{H}=\dfrac{9}{132}.100\%=6,82\%\\ \%_{P}=\dfrac{31}{132}.100\%=23,48\%\\ \%_{O}=100\%-23,48\%-6,82\%-21,21\%48,49\% \end{cases}\)

\(4,M_{C_2H_5COOCH_3}=88(g/mol)\\ \begin{cases} \%_{C}=\dfrac{48}{88}.100\%=54,55\%\\ \%_{H}=\dfrac{8}{88}.100\%=9,09\%\\ \%_{O}=100\%-9,09\%-54,55\%=36,36\% \end{cases}\)

Bài 2:

\(c,\%_{Al(AlCl_3)}=\dfrac{27}{27+35,5.3}.100\%=20,22\%\\ \%_{Al(Al_2O_3)}=\dfrac{27.2}{27.2+16.3}.100\%=52,94\%\\ \%_{Al(AlBr_3)}=\dfrac{27}{27+80.3}.100\%=10,11\%\\ \%_{Al(Al_2S_3)}=\dfrac{27.2}{27.2+32.3}.100\%=36\%\)

Vậy \(Al_2O_3\) có \(\%Al\) cao nhất và \(AlBr_3\) có \(\%Al\) nhỏ nhất

Bài 2:

\(a,\%_{Fe(FeO)}=\dfrac{56}{56+16}.100\%=77,78\%\\ \%_{Fe(Fe_2O_3)}=\dfrac{56.2}{56.2+16.3}.100\%=70\%\\ \%_{Fe(FeS)}=\dfrac{56}{56+32}.100\%=63,64\%\\ \%_{Fe(FeS_2)}=\dfrac{56}{56+32.2}=46,67\%\)

Vậy \(FeO\) có \(\%Fe\) lớn nhất và \(FeS_2\) có \(\%Fe\) thấp nhất

\(b,\%_{O(NO_2)}=\dfrac{16}{16.2+14}.100\%=69,57\%\\ \%_{O(NO)}=\dfrac{16}{16+14}.100\%=53,33\%\\ \%_{O(N_2O)}=\dfrac{16}{14.2+16}.100\%=36,36\%\\ \%_{O(N_2O_3)}=\dfrac{16.3}{14.2+16.3}.100\%=63,16\%\)

Vậy \(NO_2\) có \(\%O\) lớn nhất và \(N_2O\) có \(\%O\) nhỏ nhất

\(M_{\left(NH_4\right)_2SO_4}=\left(14+1.4\right).2+32+16.4=132\left(\dfrac{g}{mol}\right)\)

\(\%N=\dfrac{14.2}{132}.100\%=21,2\%\)

\(\%H=\dfrac{1.4.2}{132}.100\%=6\%\)

\(\%S=\dfrac{32}{132}.100\%=24,2\%\)

\(\%O=\dfrac{16.4}{132}.100\%=48,5\%\)

\(n_{H_2}=\dfrac{0,784}{22,4}=0,035\left(mol\right)\\ Đặt:n_{Al}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow\left\{{}\begin{matrix}27a+56b=1,39\\1,5a+b=0,035\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,01\\b=0,02\end{matrix}\right.\\ \Rightarrow\%m_{Al}=\dfrac{0,01.27}{1,39}.100=37,53\%\\ \Rightarrow\%m_{Fe}=100\%-37,53\%=62,47\%\)

\(M_{hc}=15.M_H=15(g/mol)\)

\(\Rightarrow n_C=\dfrac{15.80\%}{12}=1(mol);n_H=\dfrac{15.20\%}{1}=3(mol)\\ \Rightarrow CTHH:CH_3\)

Gọi CTHH của hợp chất là: \(\left(C_xH_y\right)_n\)

Ta có: \(M_{\left(C_xH_y\right)_n}=15.1=15\left(\dfrac{g}{mol}\right)\)

Ta lại có: \(\dfrac{x}{y}=\dfrac{\dfrac{80\%}{12}}{\dfrac{20\%}{1}}=\dfrac{1}{3}\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)

Vậy CTHH của hợp chất là: \(\left(CH_3\right)_n\)

Mà: \(M_{\left(CH_3\right)_n}=\left(12+1.3\right).n=15\left(\dfrac{g}{mol}\right)\)

\(\Leftrightarrow n=1\)

Vậy CTHH của hợp chất là: CH₃

M_hc = 15.2 = 30 (g/mol)

\(m_C=\dfrac{30.80}{100}=24\left(g\right)=>n_C=\dfrac{24}{12}=2\left(mol\right)\)

\(m_H=\dfrac{30.20}{100}=6\left(g\right)=>n_H=\dfrac{6}{1}=6\left(mol\right)\)

=> CTHH: C₂H₆

\(\%_{N\left(\left(NH_2\right)_2CO\right)}=\dfrac{14.2}{\left(14+2\right).2+12+16}.100\%=46,67\%\\ \%_{N\left(NH_4NO_3\right)}=\dfrac{14.2}{14.2+4+16.3}.100\%=35\%\\ \%_{N\left(NH_4Cl\right)}=\dfrac{14}{14+4+35,5}.100\%=26,17\%\\ \%_{N\left(\left(NH_4\right)H_2PO_4\right)}=\dfrac{14}{14+4+2+31+16.4}.100\%=12,17\%\)

Vậy \((NH_2)_2CO\) có hàm lượng \(N\) cao nhất

M(NH2)2CO=60g/mol

\(\Leftrightarrow\) %N[(NH2)2CO]=\(\dfrac{14.2.100}{60}\)=47%

M NH4NO3=80g/mol

\(\Leftrightarrow\) %N(NH4NO3)=\(\dfrac{14.100}{80}\)= 18%

M NH4Cl=53,5g/mol

\(\Leftrightarrow\) %N(NH4Cl)=\(\dfrac{14.100}{53.5}\)=26%

M NH4H2PO4 = 115g/mol

\(\Leftrightarrow\) %N(NH4H2PO4)=\(\dfrac{14.100}{115}\)=12%

vậy hàm lượng N trong hợp chất (NH2)2CO lớn nhất