a) 4Al + 3O2 ➜ 2Al2O3 (1)
X: Al dư và Al2O3
2Al + 6HCl ➜ 2AlCl3 + 3H2 (2)
Al2O3 + 6HCl ➜ 2AlCl3 + 3H2O (3)
\(n_{H_2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
b) Theo Pt2: \(n_{Al}dư=\frac{2}{3}n_{H_2}=\frac{2}{3}\times0,15=0,1\left(mol\right)\)
\(\Rightarrow m_{Al}dư=0,1\times27=2,7\left(g\right)\)
\(\Rightarrow m_{Al_2O_3}=12,9-2,7=10,2\left(g\right)\)
\(\Rightarrow n_{Al_2O_3}=\frac{10,2}{102}=0,1\left(mol\right)\)
Theo PT1: \(n_{Al}pư=2n_{Al_2O_3}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{Al}pư=0,2\times27=5,4\left(g\right)\)
Vậy \(m=m_{Al}pư+m_{Al}dư=5,4+2,7=8,1\left(g\right)\)
c) Theo PT1: \(n_{O_2}=\frac{3}{2}n_{Al_2O_3}=\frac{3}{2}\times0,1=0,15\left(mol\right)\)
\(\Rightarrow m_{O_2}=0,15\times32=4,8\left(g\right)\)
Theo PT2: \(n_{HCl}=3n_{Al}=3\times0,1=0,3\left(mol\right)\)
Theo PT3: \(n_{HCl}=6n_{Al_2O_3}=6\times0,1=0,6\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,3+0,6=0,9\left(mol\right)\)
\(\Rightarrow\Sigma m_{HCl}=0,9\times36,5=32,85\left(g\right)\)
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