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? Show all of the work needed to solve this problem.
2 Al (s) + 3 Fe(NO3)2 (aq) 3 Fe (s) + 2 Al(NO3)3 (aq)
Refer to the Zn problem. This is just another stoichiometry problem.
To find the number of grams of iron metal produced, we need to use stoichiometry - a method that relates the quantities of substances involved in a chemical reaction.
First, let's calculate the number of moles of iron (II) nitrate using its mass and molar mass:
Molar mass of Fe(NO3)2 = atomic mass of Fe + (2 x atomic mass of N) + (6 x atomic mass of O)
= (55.85 g/mol) + (2 x 14.01 g/mol) + (6 x 16.00 g/mol)
= 55.85 g/mol + 28.02 g/mol + 96.00 g/mol
= 179.87 g/mol
Now, we can use the molar mass and the mass percent of iron (II) nitrate to calculate the mass of iron (II) nitrate in the solution:
Mass of iron (II) nitrate = (mass percent/100) x mass of solution
= (80.5/100) x 245 g
= 0.805 x 245 g
= 197.225 g
Next, we can convert the mass of iron (II) nitrate to moles using its molar mass:
Moles of Fe(NO3)2 = Mass of iron (II) nitrate/molar mass
= 197.225 g/179.87 g/mol
= 1.095 moles
According to the balanced equation, the ratio between Al and Fe(NO3)2 is 2:3. Therefore, the number of moles of Al is:
Moles of Al = (2/3) x Moles of Fe(NO3)2
= (2/3) x 1.095 moles
= 0.73 moles
Now, we can calculate the number of moles of Fe produced using the stoichiometry ratio from the balanced equation:
Moles of Fe = (3/2) x Moles of Al
= (3/2) x 0.73 moles
= 1.095 moles
Finally, we can convert the moles of Fe to grams using its molar mass:
Mass of Fe produced = Moles of Fe x Molar mass of Fe
= 1.095 moles x 55.85 g/mol
= 61.09 grams
Therefore, you can expect approximately 61.09 grams of iron metal to be produced.