Em thảm khảo nha :
\(\begin{array}{l}
CTHH:{X_2}{O_5}\\
\% X = \dfrac{{2{M_X}}}{{2{M_X} + 5{M_O}}} \times 100\% = 43,662\% \\
{M_X} = \dfrac{{5 \times 16 \times 0,43662}}{{2 - 2 \times 0,43662}} = 31dvC\\
\Rightarrow X:Photpho(P)\\
CTHH:{P_2}{O_5}
\end{array}\)