Bài 7 :
\(m_{ct}=\dfrac{7,3.150}{100}=10,95\left(g\right)\)
\(n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)
Pt : \(2Al+6HCl\rightarrow2AlCl_3+3H_2|\)
2 6 2 3
0,1 0,3 0,1 0,15
a) \(n_{Al}=\dfrac{0,3.2}{6}=0,1\left(mol\right)\)
⇒ \(m_{Al}=0,1.27=2,7\left(g\right)\)
\(n_{H2}=\dfrac{0,3.3}{6}=0,15\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,15.22,4=3,36\left(l\right)\)
b) \(n_{AlCl3}=\dfrac{0,15.2}{3}=0,1\left(mol\right)\)
⇒ \(m_{AlCl3}=0,1.133,5=13,35\left(g\right)\)
\(m_{ddspu}=2,7+150-\left(0,15.2\right)=152,4\left(g\right)\)
\(C_{ÁlCl3}=\dfrac{13,35.100}{152,4}=8,76\)0/0
Chúc bạn học tốt
\(n_{HCl}=\dfrac{150.7,3\%}{36,5}=0,3\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ 0,1........0,3.......0,1.......0,15\left(mol\right)\\ a.m=m_{Al}=0,1.27=2,7\left(g\right)\\ V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ b.C\%_{ddAlCl_3}=\dfrac{0,1.133,5}{2,7+150-0,15.2}.100\approx8,76\%\)
