\(M_{Na_2SO_4}=23.2+32+16.4=142\left(DvC\right)\\ \%Na=\dfrac{23.2}{142}.100\%=32,4\%\\ \%S=\dfrac{32}{142}.100\%=22,5\%\\ \%O=100\%-32,4\%-22,5\%=45,1\%\)
\(-Na_2SO_4\)
\(\%_{Na}=\dfrac{23.2}{142}.100\%=32,4\%\)
\(\%_S=\dfrac{32}{142}.100\%=22,5\%\)
\(\%_O=100\%-32,4\%-22,5\%=45,1\%\)
\(-Fe_3O_4\)
\(\%_{Fe}=\dfrac{56.3}{232}.100\%=72,4\%\)
\(\%_O=100\%-72,4\%=27,6\%\)
\(-BaCl_2\)
\(\%_{Ba}=\dfrac{137}{208}.100\%=65,9\%\)
\(\%_{Cl}=100\%-65,9\%=34,1\%\)
* Na2SO4:
\(\%m_{Na}=\dfrac{NTK_{Na}.2}{NTK_{Na}.2+NTK_S+4.NTK_O}.100\%=\dfrac{23.2}{23.2+32+4.16}.100=\dfrac{46}{142}.100\approx32,394\%\\ \%m_S=\dfrac{NTK_S}{NTK_{Na}.2+NTK_S+4.NTK_O}=\dfrac{32}{142}.100\approx22,535\%\\ \Rightarrow\%m_O\approx100\%-\left(32,394\%+22,535\%\right)\approx45,071\%\)
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