Question

cho 20,4g AL2O3 hòa tan hết vào 150 g đ HCl thu ssc ALCL3 .

a) viết pthh

b) tính c% đ hcl

C) tính c% đ ALCL3

 

Answer 1

a) \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

b) \(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)

Theo PTHH: \(n_{HCl}=6.n_{Al_2O_3}=1,2\left(mol\right)\Rightarrow m_{HCl}=1,2.36,5=43,8\left(g\right)\)

=> \(C\%_{HCl}=\dfrac{43,8}{150}.100\%=29,2\%\)

c) Theo PTHH: 

\(n_{AlCl_3}=2.n_{Al_2O_3}=0,4\left(mol\right)\Rightarrow m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\)

=> \(C\%_{AlCl_3}=\dfrac{53,4}{20,4+150}.100\%=31,34\%\)