Question

Answer 1

Em làm được bài nào trong 2 bài này chưa em?

Answer 2

B2:

\(Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\\ n_{Fe}=n_{FeCl_2}=n_{H_2}=0,1\left(mol\right)\\ m_{Fe}=0,1.56=5,6\left(g\right)\\ m_{\text{dd}sau}=m_{Fe}+m_{\text{dd}HCl}-m_{H_2}=5,6+146-0,1.2=151,4\left(g\right)\\ n_{HCl}=2.0,1=0,2\left(mol\right)\\ C\%_{\text{dd}HCl}=\dfrac{0,2.36,5}{146}.100=5\%\)