Question

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Answer 1

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

\(b,n_{Al}=\dfrac{m}{M}=\dfrac{16,2}{27}=0,6\left(mol\right)\\ Theo.PTHH:n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,6=0,9\left(mol\right)\\ V_{H_2\left(đktc\right)}=n.22,4=0,9.22,4=20,16\left(l\right)\)

\(c,Theo.PTHH:n_{HCl}=3.n_{Al}=3.0,6=1,8\left(mol\right)\\ m_{HCl}=n.M=1,8.36,5=65,7\left(g\right)\)

Answer 2

a) PTHH: 2Al + 6HCl -> 2AlCl₃ + 3H₂

b) \(n_{Al}=\dfrac{16,2}{27}=0,6\left(mol\right)\\ =>n_H=0,6\left(mol\right)\\ =>V_H=0,6.22,4=13,44\left(l\right)\)

c) \(m_{Al}+m_{HCl}=m_{AlCl_3}\\ =>m_{AlCl_3}=16,2+16,2=32,4\left(g\right)\)