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 reacts with excess aluminum metal? Show all of the work needed to solve this problem.

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

Answer 1

@chmvijay

Answer 2

@sammyalabamy

Answer 3

1. you have to calculate how many moles of Fe are in the 325 g of the 87.5% by mass solution.
When you said 87.5% by mass I will assume that for every 100 grams of solution you will have 87.5 g of Iron (II) nitrate.
you can do a conversion factor "87.5g of Iron (II) nitrate = 100 grams of solution".

87.5 g iron (II) nitrate / 100 grams of solution

The other conversion factor that you need is to convert mass of Iron (II) nitrate to moles of Iron (II) nitrate. then you have to calculate the molecular mass of Iron (II) nitrate (sum the atomic mass of all the components of the Iron (II) nitrate).

1 mole of Iron (II) nitrate / molecular mass of Iron (II) nitrate

2. According to the soichiometry of the reaction 3 moles of Iron (II) nitrate will produce 3 moles of Fe you will have a conversion factor equal to 1

3 moles of Fe/3 moles of Iron (II) nitrate = 1

3. finally convert the moles of Fe to g of iron with a conversion factor similar to the one used in step 2

atomic mass of Fe/ 1 mole of Fe

put all the factors along with the initial mass of solution and you will be able to calculate the mass of Fe


325 g solution x (87.5 g iron (II) nitrate / 100 grams of solution) x (1 mole of Iron (II) nitrate / molecular mass of Iron (II) nitrate) x (3 moles of Fe/3 moles of Iron (II) nitrate ) x (atomic mass of Fe/ 1 mole of Fe) = mass of Fe