PTHH : Fe + 2HCl ---> FeCl2 + H2
a) Số mol Fe:
nFe = 5,6/56=0,1 (mol) => nHCl = 0,2 (mol).
=> mHCl = n.M = 0,2.36,5=7,3(g)
b) Số mol H2 = 0,1 (mol).
=> VH2 = n.22,4 = 0,1.22,4=2,24 (l)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\)
a)Theo PT:\(n_{HCl}=2n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,2.36,5=7,3\left(g\right)\)
b)Theo PT: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
Fe + 2HCl → FeCl2 + H2
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
a) Theo PT: \(n_{HCl}=2n_{Fe}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,2\times36,5=7,3\left(g\right)\)
b) Theo PT: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1\times22,4=2,24\left(l\right)\)
