a)
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
n Al = 2,7/27 = 0,1 mol
n HCl =200.7,3%/36,5 = 0,4(mol)
Ta thấy:
n Al / 2 = 0,05 < n HCl / 6 = 0,067 => HCl dư
n H2 = 3/2 n Al = 0,15(mol)
=> V H2 = 0,15.22,4 = 3,36 lít
b) n HCl pư = 3n Al = 0,3(mol)
=> n HCl dư = 0,4 - 0,3 = 0,1(mol)
m dd sau pư = m Al + m dd HCl - m H2 = 2,7 + 200 - 0,15.2 = 202,4 gam
Vậy :
C% AlCl3 = 0,1.133,5/202,4 .100% = 6,6%
C% HCl dư = 0,1.36,5/202,4 .100% = 1,8%
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
a) Ta có: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\n_{HCl}=\dfrac{200\cdot7,3\%}{36,5}=0,4\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{0,4}{6}\) \(\Rightarrow\) HCl còn dư
\(\Rightarrow n_{H_2}=\dfrac{3}{2}n_{Al}=0,15\left(mol\right)\) \(\Rightarrow V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\)
b) Ta có: \(\left\{{}\begin{matrix}n_{HCl\left(dư\right)}=0,1\left(mol\right)=n_{AlCl_3}\\n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Al}+m_{ddHCl}-m_{H_2}=2,7+200-0,15\cdot2=202,4\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl\left(dư\right)}=\dfrac{0,1\cdot36,5}{202,4}\cdot100\%\approx1,8\%\\C\%_{AlCl_3}=\dfrac{0,1\cdot133,5}{202,4}\cdot100\%\approx6,6\%\end{matrix}\right.\)
