Question

1) When given 6Na + Fe2O3 -> 3Na2O +2Fea. If 100.0 moles Na and 100.0 moles Fe2O3 are used in this reaction, determine the moles of solid iron produced.b. Identify the limiting reactant.c. Identify the excess reactant.d. Determine the mass from the moles of product (Fe) formed. (Hint: mole conversion)

Answer 1

According to the reactants given to us, we have the following balanced equation of the reaction:

`6Na+Fe_2O_3\rightarrow3Na_2O+2Fe`

We see that 6 moles of Na are needed to react with one mole of Fe2O3 to produce 2 moles of Fe. Now, we have 100.0 moles of Na and 100.0 moles of Fe2O3, the actual ratio Na to Fe is 1.

a) The ratio of the Na to Fe2O3 reaction is 6, as the current ratio is smaller than the theoretical ratio of the reaction, which means that we don't have enough moles of Na for the reaction, this will be the limiting reactant. So the excess reactant will be Fe2O3.

The calculations are made based on the number of moles of Na that we have, 100.0 mol. This is because Na is the limiting reactant, so the moles of iron that will be produced will be:

`\begin{gathered} MolFe=GivenmolNa*(2molFe)/(6molNa) \\ MolFe=100.0molNa*(2molFe)/(6molNa)=33.33molFe \end{gathered}`

The moles of solid iron produced are 33.33 moles of Fe.

b) As we said before, the limiting reactant will be Na

c)The excess reactant is Fe2O3

Now, to find the mass of iron, we will use the molar mass:

`\begin{gathered} gFe=GivenmolFe*(MolarMass, gFe)/(1molFe) \\ gFe=33.33molFe*(55.845gFe)/(1molFe) \\ gFe=1861gFe \end{gathered}`

d)The mass of Fe obtained will be 1861gFe