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\(M_{MgSO_4}=24+32+16.4=120\\ \%Mg=\dfrac{24}{120}.100=20\%\\ \%S=\dfrac{32}{120}.100=26,67\%\\ \%O=\dfrac{16.4}{120}.100=53,33\%\\ M_{Al\left(NO_3\right)_3}=27+62.3=213\\ \%Al=\dfrac{27}{213}.100=12,68\%\\ \%N=\dfrac{14.3}{213}.100=19,72\%\\ \%O=\dfrac{16.9}{213}.100=67,6\%\)
\(\left\{{}\begin{matrix}\%Mg=\dfrac{24.1}{120}.100\%=20\%\\\%S=\dfrac{32.1}{120}.100\%=26,667\%\\\%O=\dfrac{16.4}{120}.100\%=53,333\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%Al=\dfrac{27.1}{213}.100\%=12,676\%\\\%N=\dfrac{14.3}{213}.100\%=19,718\%\\\%O=\dfrac{16.9}{213}.100\%=67,606\%\end{matrix}\right.\)
\(\%Al=\dfrac{27}{27+62.3}.100=12,68\%\)
\(\%N=\dfrac{14.3}{27+62.3}.100=19,72\%\)
\(\%O=\dfrac{16.9}{27+62.3}.100=67,6\%\)
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:
\(a,\%_{Fe}=\dfrac{56}{180}\cdot100\%=31,11\%\\ \%_N=\dfrac{14\cdot2}{180}\cdot10\%=15,56\%\\ \%_O=100\%-31,11\%-15,56\%=53,33\%\\ b,\%_{N\left(N_2O\right)}=\dfrac{14\cdot2}{44}\cdot100\%=63,63\%\\ \%_{O\left(N_2O\right)}=100\%-63,63\%=36,37\%\\ \%_{N\left(NO\right)}=\dfrac{14}{30}\cdot100\%=46,67\%\\ \%_{O\left(NO\right)}=100\%-46,67\%=53,33\%\\ \%_{O\left(NO_2\right)}=\dfrac{16\cdot2}{46}\cdot100\%=69,57\%\\ \%_{N\left(NO_2\right)}=100\%-69,57\%=30,43\%\)
\(\%m_{Ca}=\dfrac{40}{164}.100=24,39\left(\%\right)\\ \%m_N=\dfrac{28}{164}.100=17,07\left(\%\right)\\ \%m_O=\dfrac{48.2}{164}.100=58,54\left(\%\right)\)
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Thu gọn
\(Fe\left(NO_3\right)_3:\left\{{}\begin{matrix}\%_{Fe}=\dfrac{56}{242}\cdot100\%=23,14\%\%\\\%_N=\dfrac{14\cdot3}{242}\cdot100\%=17,36\%\\\%_O=\left(100-23,14-17,36\right)\%=59,5\%\end{matrix}\right.\)
\(K_3PO_4:\left\{{}\begin{matrix}\%_K=\dfrac{39\cdot3}{212}\cdot100\%=55,19\%\\\%_P=\dfrac{31}{212}\cdot100\%=14,62\%\\\%_O=\left(100-55,19-14,62\right)\%=30,19\%\end{matrix}\right.\)
\(Ca\left(OH\right)_2:\left\{{}\begin{matrix}\%_{Ca}=\dfrac{40}{74}\cdot100\%=54,05\%\\\%_O=\dfrac{16\cdot2}{74}\cdot100\%=43,24\%\\\%_H=\left(100-54,05-43,24\right)\%=2,71\%\end{matrix}\right.\)
\(P_2O_5:\left\{{}\begin{matrix}\%_P=\dfrac{31\cdot2}{142}\cdot100\%=43,66\%\\\%_O=100\%-43,66\%=56,34\%\end{matrix}\right.\\ SiO_2:\left\{{}\begin{matrix}\%_{Si}=\dfrac{28}{60}\cdot100\%=46,67\%\\\%_O=\left(100-46,67\right)\%=53,33\%\end{matrix}\right.\\ Fe_3O_4:\left\{{}\begin{matrix}\%_{Fe}=\dfrac{56\cdot3}{232}\cdot100\%=72,41\%\\\%_O=\left(100-72,41\right)\%=27,59\%\end{matrix}\right.\)
\(MgSO_4=120\)
\(\%Mg=\dfrac{24}{120}.100\%=20\%\)
\(\%S=\dfrac{32}{120}.100\%\text{≈}26,67\%\)
\(\%O=100-\left(20+26,67\right)\text{≈}53,33\%\)