Đáp án:
$MgCO_3$
Giải thích các bước giải:
\(\begin{array}{l}
CTHH:M{g_x}{C_y}{O_z}\\
{m_{Mg}} = 28,75\% \times 84 = 24,15g\\
{n_{Mg}} = \dfrac{{24,15}}{{24}} = 1,00625\,mol\\
{m_C} = 14,2\% \times 84 = 11,928\,g\\
{n_C} = \dfrac{{11,928}}{{12}} = 0,994\,mol\\
{m_O} = 84 - 11,928 - 24,15 = 47,922g\\
{n_O} = \dfrac{{47,922}}{{16}} = 2,995125\,mol\\
x:y:z = {n_{Mg}}:{n_C}:{n_O}\\
\Rightarrow x:y:z = 1,00625:0,994:2,995125 = 1:1:3\\
\Rightarrow CTHH:MgC{O_3}
\end{array}\)