Question

hoà tan 23,2 (g) fe3o4 vào 200g dd HCl a% thu đc dd x 

a) xđ a

b) xđ C% các chất trong dd x 

Answer 1

\(n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\)

Fe3O4 + 8HCl → 2FeCl3 + FeCl2 + 4H2O

Ta có: \(n_{HCl}=8n_{Fe_3O_4}=0,8\left(mol\right)\)

=> \(a=C\%_{HCl}=\dfrac{0,8.36,5}{200}.100=14,6\%\)

\(m_{ddsaupu}=23,2+200=223,2\left(g\right)\)

Ta có : \(n_{FeCl_3}=2n_{Fe_3O_4}=0,2\left(mol\right)\)

=> \(C\%_{FeCl_3}=\dfrac{0,2.162,5}{223,2}.100=14,56\%\)

Ta có : \(n_{FeCl_2}=n_{Fe_3O_4}=0,1\left(mol\right)\)

=> \(C\%_{FeCl_2}=\dfrac{0,1.127}{223,2}.100=5,69\%\)