Câu 3:
\(a,M_{KMnO_4}=158(g/mol)\\ \%_{K}=\dfrac{39}{158}.100\%=24,68\%\\ \%_{Mn}=\dfrac{55}{158}.100\%=34,81\%\\ \%_{O}=100\%-24,68\%-34,81\%=40,51\%\)
\(M_{KClO_3}=122,5(g/mol)\\ \%_{K}=\dfrac{39}{122,5}.100\%=31,84\%\\ \%_{Cl}=\dfrac{35,5}{122,5}.100\%=28,98\%\\ \%_{O}=100\%-31,84\%-28,98\%=39,18\%\)
\(M_{Al_2O_3}=102(g/mol)\\ \%_{Al}=\dfrac{54}{102}.100\%=52,94\%\\ \%_{O}=100\%-52,94\%=47,06\%\\ M_{Na_2CO_3}=106(g/mol)\\ \%_{Na}=\dfrac{46}{106}.100\%=43,4\%\\ \%_{C}=\dfrac{12}{106}.100\%=11,32\%\\ \%_{O}=100\%-43,4\%-11,32\%=45,28\%\)
\(M_{Fe_2(SO_4)_3}=400(g/mol)\\ \%_{Fe}=\dfrac{56.2}{400}.100\%=28\%\\ \%_{S}=\dfrac{32.3}{400}.100\%=24\%\\ \%_{O}=100\%-28\%-24\%=48\%\)
\(b,\) Giả sử có 100g quặng sắt và 100g quặng nhôm
\(\Rightarrow n_{Fe_2O_3}=\dfrac{100.80\%}{160}=0,5(mol)\\ n_{Al_2O_3}=\dfrac{100.60\%}{102}=0,59(mol)\\ \Rightarrow n_{Fe}=2n_{Fe_2O_3}=1(mol)\\ n_{Al}=2n_{Al_2O_3}=1,18(mol)\\ \Rightarrow \%_{Fe}=\dfrac{1.56}{100}.100\%=56\%\\ \%_{Al}=\dfrac{1,18.27}{100}.100\%=31,86\%\)
