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DerKrebs [107]
3 years ago
9

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.
Chemistry
2 answers:
Akimi4 [234]3 years ago
8 0
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 

<span>plug in molar mass of Fe(NO3)2 and Fe and calculate</span>
Korolek [52]3 years ago
8 0
<span>1.245*55.85= 69.53 gm iron metal will be produced</span>
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Consider the following: you mix 10.0 mL of CHCl3 (d = 1.492 g/mL) and 5.0 mL of CHBr3 (d = 2.890 g/mL), giving 15.0 mL of soluti
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3 years ago
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</span>therefore 4 atoms of Fe will be left over after the reaction happens.

I hope this helps.
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2 years ago
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