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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
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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.

User AhmadMarafa
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2 Answers

6 votes
1.245*55.85= 69.53 gm iron metal will be produced
User Dimitris Baltas
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6.5k points
5 votes
265g solution x (84.5g Fe(NO3)2 / 100g solution) x (1 mole Fe(NO3)2 / __g Fe(NO3)2) x (3 moles Fe(s) / 3 moles Fe(NO3)2) x (__g Fe / mole Fe) = __g Fe

plug in molar mass of Fe(NO3)2 and Fe and calculate
User NHG
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6.6k points
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