Ta có: \(m_{HCl}=200.7,3\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
____0,2_____0,4_____0,2____0,2 (mol)
a, \(m_{Fe}=0,2.56=11,2\left(g\right)\)
\(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)
b, m dd sau pư = mFe + m dd HCl - mH2 = 11,2 + 200 - 0,2.2 = 210,8 (g)
\(\Rightarrow C\%_{FeCl_2}=\dfrac{25,4}{210,8}.100\%\approx12,05\%\)
c, \(V_A=\dfrac{210,8}{1,2}=175,67\left(ml\right)=0,175\left(l\right)\)
\(\Rightarrow C_{M_{FeCl_2}}=\dfrac{0,2}{0,175}\approx1,14\left(M\right)\)