Question

How many grams of iron metal do you expect to be produced when 524 grams of an 80.5 percent by mass iron (II) nitrate solution reacts with excess aluminum metal? Show all of the work needed to solve this problem.
2 Al (s) + 3 Fe(NO3)2 (aq) yields 3 Fe (s) + 2 Al(NO3)3 (aq) (4 points)

Answer 1

I am not sure how to solve this.

Answer 2

@aaronq

Answer 3

For any type of stoichiometry problem (except maybe when using gases at STP), you need t convert the reactants given to moles.

This varies from question to question depending on how the information is presented (e.g. molarity, molality, volume and density, mass, etc.)

Were told that we have a solution that is 80.5% iron (II) nitrate by mass, meaning for every 100 g of solution 80.5 g are iron (II) nitrate (the rest is probably just water).

We need to find the mass of only iron (II) nitrate in the 524 g solution

\(\sf mass~of ~Fe(NO_3)_2=total~mass*\%~of~Fe(NO_3)_2=524 g*0.805=419.2~g\)

Now we convert the mass to moles using the molar mass:

\(\sf moles=\dfrac{mass}{Molar ~mass}=\dfrac{419.2~g} {179.8548~ g/mol}=2.33~moles \)


Next, we relate the moles of Iron(II)Nitrate to the moles of Iron metal using their stoichiometric coefficients from the balanced reaction.

\(\sf \dfrac{moles~of~Fe(NO_3)_2}{3}=\dfrac{moles~of~ Fe}{3}\)

Since they both have the same coefficient, the moles of Fe metal are the same.

Now we convert to mass, using the same formula, \(\sf moles=\dfrac{mass}{Molar ~mass}\)

except this time you use the molar mass of Fe, not \(\sf Fe(NO_3)_2\).

\(\sf 2.33~moles=\dfrac{mass}{55.85 ~g/mol}\rightarrow mass=2.33~moles*55.85 ~g/mol=130.17~g\)

Answer 4

Oh...it wasn't that hard. The question made it seem like it was a lot more confusing.

Answer 5

haha yeah, it was pretty straight forward.

Answer 6

Thank You!