Question

Hoà tan 2,7g Al ACl+6HCl=2AlCl3+3H2 T a.Tính khối lượng HCl b.Thể tích H2 ở 25°C,1bar

Answer 1

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

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

             0,1---->0,3------------------>0,15

a) \(m_{HCl}0,3.36,5=10,95\left(g\right)\)

b) \(V_{H2\left(25^oC,1bar\right)}=0,15.24,79=3,7185\left(l\right)\)

Answer 2

\(a.n_{Al}=\dfrac{2,7}{27}=0,1mol\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{HCl}=\dfrac{0,1.6}{2}=0,3mol\\ m_{HCl}=0,3.36,5=10,59g\\ b.n_{H_2}=\dfrac{0,1.3}{2}=0,15mol\\ V_{H_2,đkc}=0,15.24,79=3,7185l\)