a;
%mFe=\(\dfrac{56.3}{56.3+16.4}.100\%=72,1\%\)
%mO=100-72,1=27,9%
b;
%mAl=\(\dfrac{27}{78}.100\%=34,6\%\)
%mO=\(\dfrac{16.3}{78}.100\%=61,54\%\)
%mH=100-34,6-61,54=3,86%
a) \(M_{Fe_3O_4}=3.56+4.16=232\)
% m Fe = \(\dfrac{3.56}{232}.100\%=72,4\%\)
% m O = 100% - 72,4% = 27,6%
b) \(M_{Al\left(OH\right)_3}=27+3.\left(16+1\right)=78\)
% m Al = \(\dfrac{27}{78}.100\%=34,6\%\)
% m O = \(\dfrac{3.16}{78}.100\%=61,5\%\)
% m H = 100% - 34,6% - 61,5% = 3,9%
a.\(M_{Fe_3O_4}=56.3+16.4=232\left(g/mol\right)\)
Trong 1mol Fe3O4 có:3 mol nguyên tử Fe:4 mol nguyên tử O
\(\%Fe=\dfrac{56.3}{232}.100\%=72,4\%\)
\(\%O=100\%-\%Fe=100\%-72,4\%=27,6\%\)
b.\(M_{Al\left(OH\right)_3}=27+\left(16+1\right).3=78\left(g/mol\right)\)
Trong 1mol Al(OH)3 có:1 mol nguyên tử Al:3 mol nguyên tử O:3 mol nguyên tử H
\(\%Al=\dfrac{27}{78}.100\%=34,6\%\)
\(\%O=\dfrac{16.3}{78}.100\%=61,5\%\)
\(\%H=100\%-\%Al-\%O=100\%-34,6\%-61,5\%=3,9\%0\)
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