Ta có :
\(n_{Al}=\frac{5,4}{27}=0,2\left(mol\right)\)
\(n_{HCl}=\frac{200.7,3\%}{36,5}=0,4\left(mol\right)\)
\(PTHH:\text{2Al+ 6HCl}\rightarrow\text{2AlCl3 +3H2}\)
b, Thấy :\(\frac{0,2}{2}>\frac{0,4}{6}\)
\(\Rightarrow\text{HCl pư ht, Al dư}\)
\(n_{H2}=\frac{3}{2}.0,4=0,6\left(mol\right)\)
\(V_{HCL}=0,6.22,4=13,44\left(l\right)\)
\(\text{c) m dd sau = 5,4+ 200 - 0,6}.\text{2= 204,2 g}\)
\(\Rightarrow C\%_{AlCl3}=\frac{0,4.\frac{2}{6}.133,5}{204,2}.100\%=8,72\%\)
a) 2Al+6HCl---->2AlCl3+3H2
b) n Al=\(\frac{5,4}{27}=0,2\left(mol\right)\)
n=\(\frac{200.7,3}{100.36,5}=0,4\left(mol\right)\)
----> Al dư
Theo pthh
n H2=\(\frac{1}{2}n_{HCl}=0,2\left(mol\right)\)---> m H2=0,4(g)
V\(_{H2}=0,2.22,4=4,48\left(l\right)\)
c) dd sau pư làAlCl3
m dd=200+5,4-0,,4=105(g)
n \(_{AlCl3}=\frac{1}{3}n_{HCl}=\frac{2}{15}\left(mol\right)\)
C%=\(\frac{\frac{2}{15}.133,5}{205}.100\%=8,68\%\)
