ta có CTHH: \(Ba^{II}_x\left(PO_4\right)^{III}_y\)
\(\rightarrow II.x=III.y\rightarrow\dfrac{x}{y}=\dfrac{III}{II}=\dfrac{3}{2}\rightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)
\(\rightarrow CTHH:Ba_3\left(PO_4\right)_2\)
Gọi CTTQ \(Ba_x\left(PO_4\right)_y\)
Theo quy tắc hóa trị
⇒ \(II.x=III.y\)
⇒ \(\dfrac{x}{y}=\dfrac{III}{II}=\dfrac{3}{2}\)
⇒ \(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)
⇒ \(CTHH:Ba_3\left(PO_4\right)_2\)
Các bước lập CTHH: a.x=b.y
CTHH:Bax(PO4)y
\(\dfrac{x}{y}=\dfrac{III}{II}=\dfrac{3}{2}\)
=> CTHH: Ba3(PO4)2
