\(n_{Fe}=\dfrac{m}{M}=\dfrac{5,6}{56}=0,1mol\)
\(m_{HCl}=\dfrac{100.15}{100}=15gam\rightarrow\)\(n_{HCl}=\dfrac{m}{M}=\dfrac{15}{36,5}\approx0,41mol\)
Fe+2HCl\(\rightarrow\)FeCl2+H2
-Tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,41}{2}\)suy ra HCl dư
\(n_{HCl\left(pu\right)}=2n_{Fe}=0,2mol\)
\(n_{HCl\left(dư\right)}=0,41-0,2=0,21mol\)
\(m_{HCl\left(dư\right)}=0,21.36,5=7,665g\)
\(n_{FeCl_2}=n_{H_2}=n_{Fe}=0,1mol\)
\(m_{FeCl_2}=0,1.127=12,7gam\)
\(V_{H_2}=0,1.22,4=2,24l\)