Đáp án:

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Giải thích các bước giải:

\(\begin{array}{l}
F{e_3}{O_4}\\
M = 232\\
\% {m_{Fe}} = \dfrac{{56 \times 3}}{{232}} \times 100\%  = 72,41\% \\
\% {m_O} = \dfrac{{16 \times 4}}{{232}} \times 100\%  = 27,59\% \\
Ca{(N{O_3})_2}\\
M = 164\\
 \to \% {m_{Ca}} = \dfrac{{40}}{{164}} \times 100\%  = 24,39\% \\
 \to \% {m_N} = \dfrac{{14 \times 2}}{{164}} \times 100\%  = 17,07\% \\
 \to \% {m_O} = \dfrac{{16 \times 6}}{{164}} \times 100\%  = 58,54\% \\
N{a_2}C{O_3}\\
M = 106\\
 \to \% {m_{Na}} = \dfrac{{23 \times 2}}{{106}} \times 100\%  = 43,40\% \\
 \to \% {m_C} = \dfrac{{12}}{{106}} \times 100\%  = 11,32\% \\
 \to \% {m_O} = \dfrac{{16 \times 3}}{{106}} \times 100\%  = 45,28\% 
\end{array}\)

$Fe_3O_4$

$\%Fe=\dfrac{3.56.100}{232}\%=72,4\%$

$\%O=\dfrac{4.16.100}{232}\%=27,6\%$

$Ca(NO_3)_2$

$\%Ca=\dfrac{40.100}{164}\%=24,4\%$

$\%N=\dfrac{14.2.100}{164}\%=17\%$

$\%O=\dfrac{16.6.100}{164}\%=58,6\%$

$NaCO_3$

$\%Na=\dfrac{23.100}{83}\%=27,7\%$

$\%C=\dfrac{12.100}{83}\%=14,4\%$

$\%O=\dfrac{16.3.100}{83}=57,9\%$