a) \(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\)
b) \(n_{H_2}=\frac{V_{H_2}}{22,4}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PTHH, ta có:
\(n_{HCl}=2n_{H_2}=2.0,15=0,3\left(mol\right)\)
\(m_{HCl}=n_{HCl}.M_{HCl}=0,3.36,5=10,95\left(g\right)\)
c) Theo PTHH, ta có:
\(n_{FeCl_2}=n_{H_2}=0,15\left(mol\right)\)
\(m_{FeCl_2}=n_{FeCl_2}.M_{FeCl_2}=0,15.127=19,05\left(g\right)\)
Giúp mink điii =^=