2Al + 6HCl → 2AlCl3 + 3H2 (1)
Mg + 2HCl → MgCl2 + H2 (2)
\(n_{HCl}=0,5\times2,5=1,25\left(mol\right)\)
Gọi \(n_{Al}=x\left(mol\right)\Rightarrow n_{Mg}=\frac{4}{3}x\left(mol\right)\)
Ta có: \(m_{Al}+m_{Mg}=8,85\)
\(\Leftrightarrow27x+24\times\frac{4}{3}x=8,85\)
\(\Leftrightarrow59x=8,85\)
\(\Leftrightarrow x=0,15\left(mol\right)\)
Vậy \(n_{Al}=0,15\left(mol\right)\) ; \(n_{Mg}=0,15\times\frac{4}{3}=0,2\left(mol\right)\)
Giả sử hh X pứ hết
Theo PT1: \(n_{HCl}pư=3n_{Al}=3\times0,15=0,45\left(mol\right)\)
Theo PT2: \(n_{HCl}pư=2n_{Mg}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}pư=0,45+0,4=0,85\left(mol\right)< 1,25\left(mol\right)\)
Vậy hh X pư hết, HCl dư
a) Theo Pt1: \(n_{H_2}=\frac{3}{2}n_{Al}=\frac{3}{2}\times0,15=0,225\left(mol\right)\)
Theo Pt2: \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow\Sigma n_{H_2}=0,2+0,225=0,425\left(mol\right)\)
\(\Rightarrow V=V_{H_2}=0,425\times22,4=9,52\left(l\right)\)
b) dung dịch Y gồm: HCl dư, AlCl3 và MgCl2
\(n_{HCl}dư=1,25-0,85=0,4\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}dư=\frac{0,4}{0,5}=0,8\left(M\right)\)
Theo PT1: \(n_{AlCl_3}=n_{Al}=0,15\left(mol\right)\)
\(\Rightarrow C_{M_{AlCl_3}}=\frac{0,15}{0,5}=0,3\left(M\right)\)
Theo PT2: \(n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow C_{M_{MgCl_2}}=\frac{0,2}{0,5}=0,4\left(M\right)\)