Question

Giúp mình với ạ

Answer 1

\(1,\)

\(n_{Fe}=\dfrac{m}{M}=\dfrac{2,8}{56}=0,05mol\)

\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

               \(0,05\) :  \(0,1\)  :  \(0,1\)   :   \(0,1\)     \(\left(mol\right)\)

\(V_{H_2}=n.22,4=0,1.22,4=2,24l\)

\(2,m_{HCl}=n.M=0,1.36,5=3,65g\)