\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{HCl}=1\cdot0,1=0,1\)
\(Fe+2HCl\xrightarrow[]{}FeCl_2+H_2\uparrow\)
0,05 ← 0,1 → 0,05
\(\Rightarrow V_{H_2}=0,05\cdot22,4=1,12\left(l\right)\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(C_{M_{HCl}}=\dfrac{n_{HCl}}{V_{HCl}}\Rightarrow n_{HCl}=C_{M_{HCl}}.V_{HCl}=1.\left(100:1000\right)=0,1\left(mol\right)\)
\(PTHH:Fe+2HCl\xrightarrow[]{}FeCl_2+H_2\)
Theo phương trình Fe dư
\(PTHH:Fe+2HCl\xrightarrow[]{}FeCl_2+H_2\)
tỉ lệ :1 2 1 1(mol)
Số mol :0,05 0,05 0,05 0,05(mol)
\(V_{H_2}=0,05.22,4=1,12\left(l\right)\)
