\(M_{Fe(OH)_3}=56+17.3=107(đvC)\\ \%_{Fe}=\dfrac{56}{107}.100\%=52,34\%\\ \%_O=\dfrac{48}{107}.100\%=44,86\%\\ \%_H=100\%-52,34\%-44,86\%=2,8\%\)

\(\left\{{}\begin{matrix}\%Fe=\dfrac{56.1}{107}.100\%=52,336\%\\\%O=\dfrac{16.3}{107}.100\%=44,86\%\\\%H=\dfrac{1.3}{107}.100\%=2,804\end{matrix}\right.\)

Câu 2:

Trong 1 mol X: \(\left\{{}\begin{matrix}n_{Ag}=\dfrac{170.63,53\%}{108}=1\left(mol\right)\\n_N=\dfrac{170.8,23\%}{14}=1\left(mol\right)\\n_O=\dfrac{170\left(100\%-63,53\%-8,23\%\right)}{16}=3\left(mol\right)\end{matrix}\right.\)

Vậy CTHH của X là \(AgNO_3\)

Câu 1:

\(a,\%_{Fe}=\dfrac{56}{180}\cdot100\%=31,11\%\\ \%_N=\dfrac{14\cdot2}{180}\cdot10\%=15,56\%\\ \%_O=100\%-31,11\%-15,56\%=53,33\%\\ b,\%_{N\left(N_2O\right)}=\dfrac{14\cdot2}{44}\cdot100\%=63,63\%\\ \%_{O\left(N_2O\right)}=100\%-63,63\%=36,37\%\\ \%_{N\left(NO\right)}=\dfrac{14}{30}\cdot100\%=46,67\%\\ \%_{O\left(NO\right)}=100\%-46,67\%=53,33\%\\ \%_{O\left(NO_2\right)}=\dfrac{16\cdot2}{46}\cdot100\%=69,57\%\\ \%_{N\left(NO_2\right)}=100\%-69,57\%=30,43\%\)

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

\(Fe\left(NO_3\right)_3:\left\{{}\begin{matrix}\%_{Fe}=\dfrac{56}{242}\cdot100\%=23,14\%\%\\\%_N=\dfrac{14\cdot3}{242}\cdot100\%=17,36\%\\\%_O=\left(100-23,14-17,36\right)\%=59,5\%\end{matrix}\right.\)

\(K_3PO_4:\left\{{}\begin{matrix}\%_K=\dfrac{39\cdot3}{212}\cdot100\%=55,19\%\\\%_P=\dfrac{31}{212}\cdot100\%=14,62\%\\\%_O=\left(100-55,19-14,62\right)\%=30,19\%\end{matrix}\right.\)

\(Ca\left(OH\right)_2:\left\{{}\begin{matrix}\%_{Ca}=\dfrac{40}{74}\cdot100\%=54,05\%\\\%_O=\dfrac{16\cdot2}{74}\cdot100\%=43,24\%\\\%_H=\left(100-54,05-43,24\right)\%=2,71\%\end{matrix}\right.\)

\(P_2O_5:\left\{{}\begin{matrix}\%_P=\dfrac{31\cdot2}{142}\cdot100\%=43,66\%\\\%_O=100\%-43,66\%=56,34\%\end{matrix}\right.\\ SiO_2:\left\{{}\begin{matrix}\%_{Si}=\dfrac{28}{60}\cdot100\%=46,67\%\\\%_O=\left(100-46,67\right)\%=53,33\%\end{matrix}\right.\\ Fe_3O_4:\left\{{}\begin{matrix}\%_{Fe}=\dfrac{56\cdot3}{232}\cdot100\%=72,41\%\\\%_O=\left(100-72,41\right)\%=27,59\%\end{matrix}\right.\)

Câu 1:

a) Al₂O₃:

Phần trăm Al trong Al₂O₃:   \(\%Al=\dfrac{27.2}{27.2+16.3}.100=52,94\%\)

Phần trăm O trong Al₂O₃:   \(\%O=100-52,94=47,06\%\)

b) C₆H₁₂O:

Phần trăm C trong  C₆H₁₂O:  \(\%C=\dfrac{12.6}{12.6+12+16}.100=72\%\)

Phần trăm H trong  C₆H₁₂O:  \(\%H=\dfrac{1.12}{12.6+12+16}.100=12\%\)

Phần trăm O trong  C₆H₁₂O :  \(\%O=100-72-12=16\%\)

Câu 2: 

\(m_H=\dfrac{5,88.34}{100}\approx2\left(g\right)\)

\(m_S=\dfrac{94,12.34}{100}=32\left(g\right)\)

\(n_H=\dfrac{m}{M}=\dfrac{2}{1}=2\left(mol\right)\)

\(n_S=\dfrac{m}{M}=\dfrac{32}{32}=1\left(mol\right)\)

⇒ CTHH của hợp chất:  H₂S

klpt : Fe2O3 là : 56.2+16.3=160(amu)

%Fe = \(\dfrac{56.2}{160}\) . 100% = 70%

%O = \(\dfrac{16.3}{160}\) . 100% = 305

Làm ơn tick cho mk

 

%Fe = (2×56×100)/(2×56+16×3) = 70%

28,A

29, B

Tham khảo 

-Ta có:MCu(OH)2=64+16×2+1×2=98g/mol-Trong 1 mol phân tử hợp chất có:1 mol nguyên tử Cu=>Cu=1×64=64g2 mol nguyên tử O=>O=2×16=32g2 mol nguyên tử H=>2×1=2gThành phần % về khối lượng của các nguyên tố:%mCu=(64/98)×100=65,31%%mO=(32/98)×100=32,65%%mH=(2/98)×100=2,04%-Ta có:MFe2(SO4)3=56×2+32×3+16×4×3=400g/mol-Trong 1 mol phân tử hợp chất có:2 mol nguyên tử Fe=>Fe=2×56=112g4 mol nguyên tử S=>S=32×4=128g12 mol nguyên tử O=>O=12×16=192gThành phần % về khối lượng của các nguyên tố:%mFe=(112/400)×100=28%%mS=(128/98)×100=32%%mO=(192/98)×100=48%   

\(M=242+132=374(đvc)\\ \%_{Fe}=\frac{56}{374}.100=14,97\%\\ \%_{N}=\frac{14.3+14.2}{374}.100=18,72\%\\ \%_H=\frac{1.4.2}{374}.100=2,14\%\\ \%_S=\frac{32}{374}.100=8,57\%\\ \%_O=55,6\%\)

\(Fe(NO_3)_3\\ \%_{Fe}=\frac{56}{242}.100\%=23,14\%\\ \%_N=\frac{14.3}{242}.100\%=17,36\%\\ \%_O=595,5\%\\ (NH_4)_2SO_4\\ \%N=\frac{14.2}{132}=21,21\% \\ \%H=\frac{(1.4).2.}{132}=6,06\% \\ \%S=\frac{32}{132}=24,24\% \\ \%O=\frac{16.4}{132}=48,49\%\\ \)