a: \(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
0,2 0,4 0,2 0,2
\(V_{H_2}=0,2\cdot22,4=4,48\left(lít\right)\)
b: \(\dfrac{n_{HCl}}{V_{HCl}}=2\)
=>\(\dfrac{0.4}{V_{HCl}}=2\)
=>\(V_{HCl}=\dfrac{0.4}{2}=0.2\left(lít\right)\)
c: \(C_M=\dfrac{n}{V}=\dfrac{0.2}{0.2}=1\)