Question

Cho mg  Al  200 ml dung dịch HCl 1,5m thu đc V(l)H2 .Tìm V,m

Answer 1

Answer 2

n_HCl=0.2 * 1.5 =0.3mol

PTHH 2Al + 6HCl -> 2AlCl₃ +3H₂

_^=>n Al= 1/3 n_HCl =0.1 mol => m _Al =0.1 * 27=2.7 g =>m=2.7

=> n_H2 =1/2 n_HCl =0.15mol =>V_H2=0.15*22.4=3.36(l)=>V=3.36

Answer 3

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

Ta có: \(n_{HCl}=1,5\cdot0,2=0,3\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,1\left(mol\right)\\n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,1\cdot27=2,7\left(g\right)\\V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\end{matrix}\right.\)

Answer 4

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

Ta có: \(n_{HCl}=0,2\cdot1,5=0,3\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,1\left(mol\right)\\n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,1\cdot27=2,7\left(g\right)\\V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\end{matrix}\right.\)