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![](https://rs.olm.vn/images/avt/0.png?1311)
ta có \(CTHH:Al_x^{III}\left(NO_3\right)_3^I\)
\(\rightarrow III.x=I.3\)
\(\Rightarrow x=\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(PTK_{Al_x\left(NO_3\right)_3}=x\cdot NTK_{Al}+3NTK_N+9NTK_O=213\\ \Rightarrow27x+3\cdot14+9\cdot16=213\\ \Rightarrow27x=27\\ \Rightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. \(Ba\left(NO_3\right)_y\xrightarrow[]{}Ba^{\left(II\right)}\left(NO_3\right)^{\left(I\right)}\)
\(\xrightarrow[]{}y=2\)
b.\(Na_xCO_3\xrightarrow[]{}Na^{\left(I\right)}CO_3^{\left(II\right)}\)
\(\xrightarrow[]{}x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
* viết dạng chung : Al\(_x\) (NO\(_3\))\(_y\)
* áp dụng qui tắt hóa trị : 3.x = 1.y
* chuyển thành tỉ lệ : \(\dfrac{x}{y}=\) \(\dfrac{1}{3}\) \(\Rightarrow\) \(\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
CTHH : Al (NO\(_3\))\(_3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta co: \(PTK_{Al_x\left(NO_3\right)_3}=213\)
\(\Rightarrow27x+14.3+16.9=213\)
\(\Rightarrow27x=27\Rightarrow x=1\)
\(\Rightarrow\) CTHH là AL(NO3)3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Al\left(NO_3\right)x=213\Leftrightarrow27+14x+48x=213\Leftrightarrow62x=213-27=186\Rightarrow x=\dfrac{186}{62}=3\)
Ta có :
$PTK = 27x + (14 + 16.3).3 = 213 \Rightarrow x = 3$