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lesya [120]
3 years ago
8

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 r

eact 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)
Chemistry
1 answer:
Greeley [361]3 years ago
6 0
The balanced reaction is:

<span>2Al (s) + 3Fe(NO3)2 (aq) = 3Fe (s) + 2Al(NO3)3 (aq)
</span>
We are given the amount of <span>iron (II) nitrate solution with its purity. This will be the starting point of our calculation.

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