How many grams of iron metal do you expect to be produced when 325 grams of an 87.5 percent by mass iron (ii) nitrate solution react with excess aluminum metals

1) Chemical reaction

Iron(II) nitrate + Al --->

3Fe(NO3)2 + 2Al ---> 2Al(NO3)3 + 3Fe

2) molar ratios

3 mol Fe(NO3)2 : 2 mol Al : 2 mol Al(NO3)3 : 3 mol Fe

3) mass of pure Fe(NO3)2

325 grams * 87.5% = 284.375

4) convert 284.375 grams into moles

number of moles = mass in grams / molar mass

molar mass Fe(NO3)2 = 55.8 g/mol + 2*14.0 g/mol + 2*3*16 g/mol = 179.8 g/mol

=> number of moles = 284.375 g / 179.8 g/mol = 1.5816 moles

4) Use proportions:

3 moles Fe(NO3)2 / 3 moles Fe = x / 1.5816 moles Fe(NO3)2

=> x = 1.5816 moles Fe

5) Convert 1.5816 moles Fe into grams

mass in grams = number of moles * atomic mass

mass in grams = 1.5816 mol * 55.8 g /mol = 88.25 grams ≈ 88.3 g (rounded to 3 significant figures)

88.3 g

The answer to this problem is. 67