Question

Giúp em với mn.Mai em thi rùi😥😥

Answer 1

a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b, \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

Theo PT: \(n_{HCl}=3n_{Al}=0,6\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,1}=6\left(M\right)\)

c, \(m_{AlCl_3\left(5\%\right)}=50.5\%=2,5\left(g\right)\)

\(m_{AlCl_3\left(20\%\right)}=50.20\%=10\left(g\right)\)

\(\Rightarrow C\%_{AlCl_3}=\dfrac{2,5+10}{50+50}.100\%=12,5\%\)