a, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
b, Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c, Theo PT: \(n_{HCl}=2n_{Fe}=0,4\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(1mol\) \(2mol\) \(1mol\)
\(0,2mol\) \(0,4mol\) \(0,2mol\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(V_{H_2}=n.22,4=0,2.22,4=4,48\left(l\right)\)
\(m_{HCl}=n.M=0,4.=14,6\left(g\right)\)