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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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We have to find the amount of AgCl formed from 100 g of Silver nitrate by writing the expression.

100 g \text { of } A g N O_{3} \times \frac{2 \text { mol } A g N O_{3}}{169.87 g A g N O_{3}} \times \frac{2 \text { mol } A g C l}{1 \text { mol } A g N O_{3}} \times \frac{143.32 g A g C l}{1 \text { mol } A g C l}

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