Question

Đề dùng 5,4g Al cho Hdd HCl vừa đủ. a) tính vH2 thu. Đc (đktc). b) tính khối lượng HCl phản ứng c) AlCl3 sinh ra

Answer 1

2Al+6HCl->2AlCl₃+3H₂

0,2-----0,6------0,2-----0,3 mol

n_Al=\(\dfrac{5,4}{27}\)=0,2 mol

=>V_H2=0,3.22,4=6,72l

=>m _HCl=0,6.36,5=21,9g

=>m _AlCl3=0,2.133,5=26,7g

 

Answer 2

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ 2Al+6HCl\rightarrow\left(t^o\right)2AlCl_3+3H_2\\ n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{6}{2}.0,2=0,6\left(mol\right)\\ n_{H_2}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ b,m_{HCl}=0,6.36,5=21,9\left(g\right)\\ c,m_{AlCl_3}=133,5.0,2=26,7\left(g\right)\)