Cho 7,8 gam bột Zn vào 200 ml dung dịch AgNO3 1,0M. Sau khi phản...

${{n}_{Zn}}=\dfrac{7,8}{65}=0,12 mol; {{n}_{AgN{{O}_{3}}}}=0,2 mol$
$Zn+AgN{{O}_{3}}\xrightarrow{{}}\left| \begin{aligned}
& Dung d\grave{o}ch Zn{{(N{{O}_{3}})}_{2}}:0,1 mol (BT.N) \\
& m gam\left\{ \begin{aligned}
& Ag:0,12 (BT.Ag) \\
& Zn d\ddot{o}:0,02 mol (BT.Zn) \\
\end{aligned} \right. \\
\end{aligned} \right.$
$\to m=108.0,12+65.0,02=14,26 gam$
Note 6: Bài toán bảo toàn liên kết $\pi $
- Bài toán: $Ho\tilde{a}n h\hat{o}\ddot{i}p X\left\{ \begin{aligned}
& hi\tilde{n}rocacbon \\
& {{H}_{2}} \\
\end{aligned} \right.\xrightarrow[t{}^\circ ]{Ni}Y\xrightarrow{+dd B{{r}_{2}}}\left\langle \begin{aligned}
& Z (kho\hat{a}ng pha\hat{u}n \ddot{o}\grave{u}ng v\hat{o}\grave{u}i B{{r}_{2}}) \\
& {{m}_{dd B{{r}_{2}} ta\hat{e}ng}} \\
\end{aligned} \right.$
$\xrightarrow{BTKL}\left\{ \begin{aligned}
& \to \to \\
& \\
\end{aligned} \right.$
$\xrightarrow{Ba\hat{u}o toan mol \pi }{{n}_{B{{r}_{2}} p\ddot{o}}}+{{n}_{{{H}_{2}} p\ddot{o}}}={{n}_{\pi }}={{n}_{anken}}+2{{n}_{ankin}}+3{{n}_{vinylaxetilen}}+...$
$X\xrightarrow{Ni,t{}^\circ }Y$ ta thấy số mol khí giảm bằng ${{n}_{{{H}_{2}} p\ddot{o}}}\to $