How many grams of iron metal do you expect to be produced when 245 grams of an 80.5 percent by mass iron (II) nitrate solution reacts with excess aluminum metal?

Question

abmon98 · Answer

3Fe(NO3)^2+2Al-->2Al(NO3)^3+3Fe
245*80.5/100=197 g Fe(NO3)2  out of the 245 grams of solution.
Use Number of Moles=Mass(g)/Molar Mass(g/mol)
197/(55.8+(14*2)+(16*6))
197/179.8=1.09 moles of 
Look at the mole ratio of Fe(NO3)^2 to Fe its 3:3 when simplified 1:1 they share the same number of moles.
Atomic weight of Fe is 55.8
Number of moles*molar mass=Mass
1.10*55.8=61.1 grams