What mass of zinc metal will be oxidized if #15.0*g# of #FeI_3# react to give iron metal?

Question

Luna Chen · Accepted Answer

Approx. #4*g# zinc will be oxidized

Zinc is higher on the activity series than iron; it should therefore reduce the iron salt: i.e. #stackrel(0)Zn(s) rarr Zn^(2+) + 2e^(-)# #(i)# And for every oxidation there is a corresponding reduction.......... i.e. #Fe^(3+) + 3e^(-)rarrstackrel(0)Fe(s)# #(ii)# We cross multiply: #2xx(ii) + 3xx(i):# #3Zn(s) +2Fe^(3+) rarr 3Zn^(2+) + 2Fe(s)# Both mass and charge are conserved, so this is a valid representation.......... We has #(15.0*g)/(399.88*g*mol^-1)=0.0375*mol# ferric salt.........and given the stoichiometry this will oxidize.......... #3/2xx0.0375*molxx65.39*g*mol^-1=??*g# zinc metal.......

Jose Ayala · Answer

undefined The molar mass of FeI₃ is approximately 434.55 g/mol. The balanced equation is:

$$2FeI_3 + 3Zn \rightarrow 2Fe + 3ZnI_2$$

The stoichiometric ratio is 2 moles of FeI₃ to 3 moles of Zn. Calculate the moles of FeI₃ and then determine the moles of Zn needed. Finally, convert moles of Zn to grams using its molar mass.

$$ \text{Moles of FeI₃} = \frac{15.0\,g}{434.55\,g/mol} $$

$$ \text{Moles of Zn} = \frac{3}{2} \times \text{Moles of FeI₃} $$

$$ \text{Mass of Zn} = \text{Moles of Zn} \times \text{Molar mass of Zn} $$