\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{Fe}=0,2mol;n_{HCl}=0,6mol\)
C/m sau pư: Fe hết, HCl dư
\(\Rightarrow V_{H_2}=4,48l\)
\(m_{ddHCl}=200.1,12=224g\)
\(m_{ddsaupư}=11,2+224-0,2.2=234,8g\)
\(C\%ddHCl=\dfrac{0,2.36,5.100}{234,8}\approx3,1\%\)
\(C\%ddFeCl_2=\dfrac{0,2.127.100}{234,8}\approx10,81\%\)