Question

PLEASE HELP :)
How many grams of iron metal do you expect to be produced when 265 grams of an 84.5 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) 3 Fe (s) + 2 Al(NO3)2 (aq)

Answer 1

This is a chemistry question I'm afraid.
What you need is mole concept, which compares the number of atoms.

An 84.5% by mass Fe(II) Nitrate solution means the whole thing with water and whatever weights 265g, but 84.5% of the 265g belongs to Fe(II) Nitrate, so keep that in mind.

The next step is to reduce that mass into how many atoms of Fe(II) Nitrate there are. This can by dividing the mass (in grams) by the Mr, the molar mass, which is usually given in g/mol.

So we take grams, divide by molar mass, and the units of the answer is mol. Wonderful. We can now compare how many of this to that in number of atoms.

I notice that for every 3 of Fe(NO3)2, I get 3 Fe precipitated. This tells me that Fe(II) Nitrate produces Fe(s) in a 1:1 ratio. So The number of mole of Iron(II) Nitrate is the number of moles of Fe solid produced, assuming that the aluminum is in excess.

With all this in place, we can finally start calculating:
\[Mass\:of\:Fe(NO_{3})_{2}:\:84.5/100*265=223.925g\]\[M_{r}\:of\:Fe(NO_{3})_{2}:\:Fe+2(NO_{3})=56+2(14+16*3)=180g\:mol^{-1}\]\[Moles\:of\:Fe(NO_{3})_{2}=\frac{223.925}{180}=1.244mol\]\[Moles\:of\:Fe(NO_{3})_{2}(aq):Fe(s)\:=>3:3=>1:1\]Thus,\[Moles\:of\:Fe(s)=\frac{1}{1}*1.244=1.244mol\]
And
\[Mass\:of\:Fe(s)=1.244*56=70g\]