\(n_{Fe_3O_4}=\dfrac{8}{232}=\dfrac{1}{29}\left(mol\right)\)
\(Fe_3O_4+8HCl\rightarrow FeCl_2+2FeCl_3+4H_2O\)
\(\dfrac{1}{29}.......\dfrac{8}{29}................\dfrac{2}{29}\)
\(m_{HCl}=\dfrac{8}{29}\cdot36.5=10.06\left(g\right)\)
\(m_{FeCl_3}=\dfrac{2}{29}\cdot162.5=11.21\left(g\right)\)
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