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How many grams of iron metal do you expect to be produced when 265 grams of an 84.5 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) yields 3Fe (s) + 2Al(NO3)3 (aq)

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The balanced reaction is:

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

We are given the amount of iron (II) nitrate solution with its purity. This will be the starting point of our calculation.

265 (.845) gram iron (II) nitrate (1 mol Fe(NO3)2 / 179.85 g Fe(NO3)2) ( 3 mol Fe / 3 mol Fe(NO3)2) (55.85 g Fe / 1 mol Fe) = 69.54 g Fe

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