Tính % khối lượng của oxi trong các hợp chất sau : a. KClo3 b. KMnO4 c. KNO3 d. H2O e. Fe2O3 f. NaNo3

Em tham khảo nha :

\(\begin{array}{l}
a)\\
CTHH:KCl{O_3}\\
{M_{KCl{O_3}}} = 39 + 35,5 + 16 \times 3 = 122,5dvC\\
\% O = \dfrac{{3{M_O}}}{{{M_{KCl{O_3}}}}} \times 100\%  = \dfrac{{3 \times 16}}{{122,5}} \times 100\%  = 39,18\% \\
b)\\
CTHH:KMn{O_4}\\
{M_{KMn{O_4}}} = 39 + 55 + 16 \times 4 = 158dvC\\
\% O = \dfrac{{4{M_O}}}{{{M_{KMn{O_4}}}}} \times 100\%  = \dfrac{{4 \times 16}}{{158}} \times 100\%  = 40,5\% \\
c)\\
CTHH:KN{O_3}\\
{M_{KN{O_3}}} = 39 + 14 + 16 \times 3 = 101dvC\\
\% O = \dfrac{{3{M_O}}}{{{M_{KN{O_3}}}}} \times 100\%  = \dfrac{{3 \times 16}}{{101}} \times 100\%  = 47,52\% \\
d)\\
CTHH:{H_2}O\\
{M_{{H_2}O}} = 2 + 16 = 18dvC\\
\% O = \dfrac{{{M_O}}}{{{M_{{H_2}O}}}} \times 1005 = \dfrac{{16}}{{18}} \times 100\%  = 88,89\% \\
e)\\
CTHH:F{e_2}{O_3}\\
{M_{F{e_2}{O_3}}} = 2 \times 56 + 3 \times 16 = 160dvC\\
\% O = \dfrac{{3{M_O}}}{{{M_{F{e_2}{O_3}}}}} \times 100\%  = \dfrac{{3 \times 16}}{{160}} \times 100\%  = 30\% \\
f)\\
CTHH:NaN{O_3}\\
{M_{NaN{O_3}}} = 23 + 14 + 16 \times 3 = 85dvC\\
\% O = \dfrac{{3{M_O}}}{{{M_{NaN{O_3}}}}} \times 100\%  = \dfrac{{3 \times 16}}{{85}} \times 100\%  = 56,5\% 
\end{array}\)