Question

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 metal? Show all of the work needed to solve this problem. 2Al (s) + 3Fe(NO3)2 (aq) 3Fe (s) + 2Al(NO3)3 (aq) Using Stoichiometry please

Answer 1

Answer: 88.2 g

Solution:

1) Chemical equation:

2Al (s) + 3Fe(NO3)2 (aq) → 3Fe (s) + 2Al(NO3)3 (aq)

2) Theoretical molar ratios

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

3) Starting mass of pure iron nitrate

% = (mass of iron nitrate / mass of solution) * 100 = 87.5

=> mass of iron nitrate = 87.5 * mass of solution / 100

mass of solution = 325 g

=> mass of iron nitrate = 87.5 * 325 g / 100 = 284.375 g

4) moles of iron nitrate

moles = mass in grams / molar mass

molar mass of Fe(NO3)2 = 179.85 g/mol

moles = 284.375 g/ 179.85 g/mol = 1.58 moles Fe(NO3)2

5) proportion:

x 3 mol Fe
--------------------------- = ----------------------
1.58 mol Fe(NO3)2 3 mol Fe(NO3)2

Clear x:

x = 1.58 mol Fe

6) Convert 1.58 mol Fe into grams

mass = number of moles * atomic mass

atomic mass of iron = 55.845 g / mol

mass = 1.58 moles * 55.845 g/mol = 88.24 g

Rounded to 3 significant figures: 88.2 grams of Fe.